Example Question - kinematic equation
Here are examples of questions we've helped users solve.
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.
Solving Physics Problem: Finding Time of Free Fall
The image shows two physics problems. We'll solve the second problem, as it is the first complete one seen in the image: "2. A freely falling body starting from rest attains a velocity of 45 m/s. How long has it been falling?" To solve this question, we can use the kinematic equation for uniformly accelerated motion: v = u + at where: v = final velocity (45 m/s) u = initial velocity (0 m/s, since it starts from rest) a = acceleration (9.8 m/s², the acceleration due to gravity) t = time (this is what we're solving for) Let's plug in the known values: 45 m/s = 0 m/s + (9.8 m/s²) * t Now, solve for t: t = 45 m/s / 9.8 m/s² t ≈ 4.59 seconds So, the body has been falling for approximately 4.59 seconds.
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.
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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