Example Question - acceleration due to gravity
Here are examples of questions we've helped users solve.
Calculating the Time for a Ball to Reach Zero Speed when Thrown Upward
<p>To calculate the time for the ball to reach zero speed, we can use the following kinematic equation for uniformly accelerated motion:</p> \[ v_f = v_i + a \cdot t \] <p>Where:</p> <p>\( v_f \) is the final velocity (0 m/s when the ball reaches its maximum height and stops momentarily)</p> <p>\( v_i \) is the initial velocity (4.5 m/s)</p> <p>\( a \) is the acceleration (acceleration due to gravity; for a ball thrown upward this is -9.8 m/s\(^2\), since it acts downward)</p> <p>\( t \) is the time in seconds</p> <p>We can set \( v_f \) to 0 and solve for \( t \):</p> \[ 0 = 4.5 \text{ m/s} - (9.8 \text{ m/s}^2) \cdot t \] \[ 9.8 \text{ m/s}^2 \cdot t = 4.5 \text{ m/s} \] \[ t = \frac{4.5 \text{ m/s}}{9.8 \text{ m/s}^2} \] \[ t \approx 0.459 \text{ s} \] <p>So, the ball will take approximately 0.459 seconds to reach zero speed.</p>
Calculating the Final Velocity of a Falling Object
Para resolver la pregunta, primero necesitamos comprender qué información se nos está proporcionando y qué se nos está pidiendo que hagamos. La imagen muestra una tabla con dos filas. La primera fila indica el tiempo (en segundos) y la segunda fila indica la velocidad de una pelota (en metros por segundo) en los tiempos respectivos de 1 a 5 segundos. Tiempo (s) | 1 | 2 | 3 | 4 | 5 Velocidad (m/s) | 9.8 | 19.6 | 29.4 | 39.2 | 49 Se nos informa que la pelota tarda diez segundos en caer al suelo y se observa que la velocidad aumenta linealmente con el tiempo. Esto es característico de un objeto en caída libre bajo la influencia de la gravedad en la Tierra, donde la aceleración debida a la gravedad es aproximadamente 9.8 m/s². El incremento de velocidad entre cada segundo es de 9.8 m/s (19.6 - 9.8 = 9.8, 29.4 - 19.6 = 9.8, etc.). Esto significa que la velocidad de la pelota aumenta en 9.8 m/s cada segundo. Dado que la aceleración (a) = 9.8 m/s² y el tiempo (t) para el cual necesitamos calcular la velocidad es t = 10 s, podemos usar la fórmula de la velocidad final (v) para un objeto en caída libre: \[ v = u + at \] donde \( u \) es la velocidad inicial (en este caso, 0, ya que la pelota se deja caer y no se lanza), y \( a \) es la aceleración debida a la gravedad. Reemplazamos las variables conocidas: \[ v = 0 + (9.8 \text{ m/s}^2)(10 \text{ s}) \] \[ v = 98 \text{ m/s} \] Por lo tanto, la velocidad de la pelota después de diez segundos de caída es 98 metros por segundo.
Analysis of Velocity in Free Fall
La imagen muestra un problema de física que habla sobre el movimiento de una pelota que se deja caer y describe cómo la velocidad de la pelota cambia con respecto al tiempo. La tabla proporcionada muestra los valores de la velocidad en metros por segundo (m/s) de la pelota en diferentes momentos en segundos (s). La tabla es como sigue: Tiempo (s) | Velocidad (m/s) ------------|----------------- 1 | 9.8 2 | 19.6 3 | 29.4 4 | 39.2 5 | 49 A partir de los datos de la tabla, podemos observar que la velocidad de la pelota aumenta linealmente con el tiempo. Esto es característico de un objeto en caída libre bajo gravedad constante sin resistencia del aire. La aceleración constante debida a la gravedad en la superficie de la Tierra es aproximadamente \(9.8 \text{ m/s}^2\). La pregunta no está completamente visible en la imagen, así que asumiré que se necesita encontrar una relación o patrón en los datos. La relación que podemos deducir es que la velocidad incrementa \(9.8 \text{ m/s}\) por cada segundo que pasa. Esto se debe a que cada segundo, bajo la gravedad de la Tierra, el objeto en caída libre gana \(9.8 \text{ m/s}\) de velocidad. Esto lo podemos confirmar con los datos proporcionados donde a los 1 segundo la velocidad es \(9.8 \text{ m/s}\), a los 2 segundos el doble \(19.6 \text{ m/s}\), y así sucesivamente, aumentando \(9.8 \text{ m/s}\) por cada segundo adicional. La fórmula para calcular la velocidad (v) en cualquier tiempo (t) durante una caída libre es \(v = g \cdot t\), donde \(g\) es la aceleración debida a la gravedad.
Physics Kinematics Problem Solving
The question you're asking to solve cannot be fully seen in the provided image. It seems to be a multi-part question, but the text is cut off, especially the part with the actual question you want solved. The visible parts refer to three different problems related to physics, particularly motion under gravity: 4. A problem about a ball dropped and attaining a certain velocity in a given time interval (acceleration due to gravity is asked for). 5. A problem involving a ball thrown and its time of flight, aiming to find the height of a flagpole. 6. A problem about a stone released from a balloon that is ascending and calculations related to the time it takes for the stone to hit the ground. Since problem #4's full question is visible, I'll provide a solution for that: The question states: "If a ball is dropped and attains a velocity of 29.31 m/s in 3.00 s, what is the acceleration due to gravity?" To find the acceleration, we can use the following equation from kinematics that relates initial velocity (v_i), acceleration (a), and time (t): v = v_i + a*t Since the ball is dropped, the initial velocity v_i = 0, and the equation simplifies to: v = a*t Plugging in the values provided: 29.31 m/s = a * 3.00 s Now solve for the acceleration a: a = 29.31 m/s / 3.00 s a ≈ 9.77 m/s² This is the acceleration due to gravity, which is approximately the standard gravity acceleration on Earth's surface (9.81 m/s²). The slight difference can be attributed to rounding or measurement error.
Stone Descent Calculation
The question refers to a stone being released from a balloon that is ascending at a rate of 5.0 m/s when the balloon is at a height of 535 m above the ground. We need to determine the time it takes for the stone to reach the ground. First, we acknowledge that when the stone is released, it has the same upward velocity as the balloon, which is 5.0 m/s upwards. However, gravity will begin to act on the stone immediately after it is released, causing the stone to decelerate at a rate of approximately 9.81 m/s² (the acceleration due to gravity on Earth). We can use the kinematic equation for uniformly accelerated motion to find the time it takes for the stone to reach the ground. The equation is given by: \[ s = ut + \frac{1}{2}at^2 \] where: - \( s \) is the displacement (535 m in this case, but since the stone will travel upward before reversing direction, we will need to consider this in our calculations), - \( u \) is the initial velocity (5.0 m/s upward), - \( a \) is the acceleration due to gravity (-9.81 m/s², negative because it acts downward), - \( t \) is the time in seconds. However, we first need to calculate how much time it takes for the stone to come to a stop before it starts to fall back to the ground due to gravity. We use the following kinematic equation for that: \[ v = u + at \] Setting \( v = 0 \) since the stone comes to a stop momentarily when the initial upward speed is overcome by gravity, we get: \[ 0 = 5.0 m/s - (9.81 m/s² \times t) \] Now, solving for \( t \), which is the time for the stone to reach the peak of its trajectory: \[ t = \frac{5.0 m/s}{9.81 m/s²} \approx 0.51 s \] During this time, the stone will have traveled further upward. We calculate the upward distance using the first equation we mentioned: \[ s = ut + \frac{1}{2}at^2 \] \[ s = (5.0 m/s \times 0.51 s) + \frac{1}{2}(-9.81 m/s² \times (0.51 s)^2) \] Plugging in the numbers we have: \[ s \approx 5.0 m/s \times 0.51 s + \frac{1}{2} \times -9.81 m/s² \times 0.26 s^2 \] \[ s \approx 2.55 m - 1.27 m \] \[ s \approx 1.28 m \] The total height from which the stone will now fall to the ground is the height of the balloon plus this additional ascent: \[ H = 535 m + 1.28 m \] \[ H \approx 536.28 m \] Now, using the initial velocity as 0 m/s (since we are considering the point from which the stone starts falling down), and the total height from which it falls as \( H \), the kinematic equation becomes: \[ 536.28 m = 0 m/s \times t + \frac{1}{2}(-9.81 m/s²)t^2 \] This simplifies to: \[ 536.28 m = -4.905 t^2 \] Solving for \( t^2 \): \[ t^2 = \frac{536.28 m}{4.905} \approx 109.3 \] Now, solving for \( t \) (the time it takes for the stone to fall): \[ t \approx \sqrt{109.3} \] \[ t \approx 10.46 s \] Finally, we sum the time it took for the stone to stop its ascent (0.51 s) and the time it took to fall to the ground (10.46 s): Total time \( T = 0.51 s + 10.46 s \approx 10.97 s \) Therefore, it will take approximately 10.97 seconds for the stone to reach the ground.
Calculating Maximum Height of a Ball Thrown Upwards
The image contains a handwritten question which states: "A ball is thrown from the ground to the top of a school flagpole. If it returns to the ground after 4s. What is the height?" To solve this problem, we can use the kinematic equation for the position (height in this case) as a function of time while an object is in free fall under the influence of gravity. The equation is: \[ h = v_i t + \frac{1}{2} a t^2 \] where: - \( h \) is the height, - \( v_i \) is the initial velocity, - \( t \) is the time, - \( a \) is the acceleration due to gravity (which is approximately 9.8 m/s² downwards). Given that the round trip takes 4 seconds, the ball takes 2 seconds to reach the maximum height (since the time going up is the same as the time going down for a symmetric trajectory under gravity). Assuming that the ball is thrown and caught at the same height (ground level) and neglecting air resistance, we can take the acceleration due to gravity as -9.8 m/s² (negative because it's acting downwards). Let's set the time for half the trip (just to reach the peak) as: \[ t = 2 \text{ seconds} \] And since the ball stops at the peak (the top of its trajectory), the initial velocity can be found by using the equation (at the peak the final velocity is 0 m/s): \[ v_f = v_i + at \] \[ 0 = v_i - 9.8 \times 2 \] \[ v_i = 19.6 \text{ m/s} \] Now we can calculate the height \( h \) using the initial velocity \( v_i \) and half the total time (up trip only): \[ h = v_i t + \frac{1}{2} a t^2 \] \[ h = 19.6 \times 2 + \frac{1}{2} (-9.8) \times 2^2 \] \[ h = 39.2 - 19.6 \] \[ h = 19.6 \text{ meters} \] Therefore, the height of the school flagpole, or the maximum height achieved by the ball, is 19.6 meters.
Solving for Acceleration Due to Gravity in Physics Problem
The image shows a physics problem that reads: "If a ball is dropped and attains a velocity of 29.4 m/s in 3.00 s, what is the acceleration due to gravity?" To solve for the acceleration due to gravity, we can use the kinematic equation that relates initial velocity (v_i), final velocity (v_f), acceleration (a), and time (t): \[ v_f = v_i + at \] For an object that is dropped, the initial velocity is zero (\( v_i = 0 \)), so the equation simplifies to: \[ v_f = at \] We can now solve for acceleration (a) using the final velocity (v_f) and time (t) provided: \[ a = \frac{v_f}{t} \] Plugging in the given values: \[ a = \frac{29.4 \text{ m/s}}{3.00 \text{ s}} \] \[ a = 9.8 \text{ m/s}^2 \] So, the acceleration due to gravity is \( 9.8 \text{ m/s}^2 \), which is consistent with the average acceleration due to gravity on Earth.
Calculating Acceleration Due to Gravity from Velocity and Time
The photo shows three questions, of which the first one (#4) is: If a ball is dropped and attains a velocity of 29.4 m/s in 3.00 s, what is the acceleration due to gravity? To solve this, you can use the following kinematic equation for uniformly accelerated motion: \[ v = u + at \] where: - \( v \) is the final velocity - \( u \) is the initial velocity (which is 0 m/s, since the ball is dropped) - \( a \) is the acceleration (which we are trying to find, and we expect it to be the acceleration due to gravity) - \( t \) is the time Given that the initial velocity \( u = 0 \) m/s, the final velocity \( v = 29.4 \) m/s, and the time \( t = 3.00 \) s, you can plug these values into the equation to find \( a \): \[ 29.4 = 0 + a \times 3.00 \] Now solve for \( a \): \[ a = \frac{29.4}{3.00} \] \[ a \approx 9.8 \, \text{m/s}^2 \] Thus, the acceleration due to gravity, \( g \), is approximately 9.8 m/s², which is in fact the standard value for gravitational acceleration near the Earth's surface.
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