Example Question - projectile motion
Here are examples of questions we've helped users solve.
Determining the Time and Maximum Height of a Projectiles Launch
<p>To determine the time at which the projectile reaches its maximum height, we need to find when the velocity is zero. The velocity is the derivative of the position function \( y(t) \).</p> <p>The position function given in the question is:</p> <p>\[ y(t) = -16t^2 + 64t + 80 \]</p> <p>To find the velocity, we differentiate \( y(t) \) with respect to \( t \):</p> <p>\[ v(t) = y'(t) = \frac{d}{dt}(-16t^2 + 64t + 80) = -32t + 64 \]</p> <p>We set the velocity to zero and solve for \( t \):</p> <p>\[ 0 = -32t + 64 \]</p> <p>\[ 32t = 64 \]</p> <p>\[ t = 2 \]</p> <p>The projectile reaches its maximum height at \( t = 2 \) seconds.</p> <p>Now we calculate the maximum height by substituting \( t = 2 \) back into the position function \( y(t) \):</p> <p>\[ y(2) = -16(2)^2 + 64(2) + 80 \]</p> <p>\[ y(2) = -16(4) + 128 + 80 \]</p> <p>\[ y(2) = -64 + 128 + 80 \]</p> <p>\[ y(2) = 144 \]</p> <p>Therefore, the maximum height reached by the projectile is 144 feet.</p>
Calculating Time to Reach Zero Speed and Return Time for a Ball Thrown Upward
The first part of the problem asks for the time it will take for the ball to reach zero speed when thrown upward at a speed of \(4.5 \ ms^{-1}\). The motion of the ball can be described by the kinematic equation which includes the initial velocity (\(v_i\)), the final velocity (\(v_f\)), the acceleration due to gravity (\(g\)), and the time (\(t\)): \[v_f = v_i + at\] For the ball to come to a stop at the highest point of its trajectory, its final velocity, \(v_f\), would be \(0 \ ms^{-1}\), and the acceleration, \(a\), would be \(-9.8 \ ms^{-2}\) as it's going against gravity. By substituting these values and the given initial velocity \(v_i = 4.5 \ ms^{-1}\) into the equation above: <p>\(0 = 4.5 \ ms^{-1} - (9.8 \ ms^{-2}) \times t \)</p> <p>\(9.8 \ ms^{-2} \times t = 4.5 \ ms^{-1} \)</p> <p>\(t = \frac{4.5 \ ms^{-1}}{9.8 \ ms^{-2}} \)</p> <p>\(t \approx 0.459 \ s\)</p> This is the time it takes for the ball to reach the highest point where its speed is zero. The second part of the problem asks for the time it will take for the ball to return to its starting point. This can be calculated by considering the symmetry of projectile motion. The time taken to ascend to the highest point is the same as the time taken to descent back to the starting point. Therefore, the total time for the whole journey is double the time calculated for the upward journey: <p>\(Total\ time = 2 \times \text{time to reach maximum height} \)</p> <p>\(Total\ time = 2 \times 0.459 \ s \)</p> <p>\(Total\ time \approx 0.918 \ s \)</p> The third part of the problem asks for the final speed of the ball when it returns to its starting point. The speed of the ball when it returns to the starting point will be the same as the speed it had when it was thrown upward, but in the opposite direction (assuming no air resistance). Therefore: <p>\(Final\ speed = 4.5 \ ms^{-1}\)</p>
Projectile Motion Calculations
// Problema 1 a) Magnitud de la velocidad a los 4 segundos: <p>$$ v = v_0 + g \cdot t = 6 \, m/s + (9.8 \, m/s^2)(4 \, s) = 45.2 \, m/s $$</p> b) Distancia recorrida entre los segundos 4 y 5: <p>$$ d = v_0 \cdot t + \frac{1}{2} g \cdot t^2 $$</p> <p>$$ d_{4s} = (6 \, m/s)(4 \, s) + \frac{1}{2}(9.8 \, m/s^2)(4 \, s)^2 = 24 \, m + 78.4 \, m = 102.4 \, m $$</p> <p>$$ d_{5s} = (6 \, m/s)(5 \, s) + \frac{1}{2}(9.8 \, m/s^2)(5 \, s)^2 = 30 \, m + 122.5 \, m = 152.5 \, m $$</p> <p>$$ d_{4\_5s} = d_{5s} - d_{4s} = 152.5 \, m - 102.4 \, m = 50.1 \, m $$</p> // Problema 2 a) Distancia recorrida a los 3 segundos: <p>$$ d = v_0 \cdot t - \frac{1}{2} g \cdot t^2 = 30 \, m/s \cdot 3 \, s - \frac{1}{2}(9.8 \, m/s^2)(3 \, s)^2 = 90 \, m - 44.1 \, m = 45.9 \, m $$</p> b) Magnitud de la velocidad a los 3 segundos: <p>$$ v = v_0 - g \cdot t = 30 \, m/s - (9.8 \, m/s^2)(3 \, s) = 30 \, m/s - 29.4 \, m/s = 0.6 \, m/s $$</p> c) Altura máxima alcanzada (cuando \( v = 0 \)): <p>$$ 0 = v_0^2 - 2 g \cdot d_{max} $$</p> <p>$$ d_{max} = \frac{v_0^2}{2g} = \frac{(30 \, m/s)^2}{2(9.8 \, m/s^2)} = \frac{900}{19.6} \approx 45.9 \, m $$</p> d) El tiempo que tardará en el aire (tiempo hasta subir y bajar): <p>$$ t_{subida} = \frac{v_0}{g} = \frac{30 \, m/s}{9.8 \, m/s^2} \approx 3.06 \, s $$</p> <p>$$ t_{total} = 2 \cdot t_{subida} \approx 2 \cdot 3.06 \, s = 6.12 \, s $$</p>
Calculating the Maximum Height Reached by a Ball Thrown Vertically Upward
<p>To find the maximum height \( h \) reached by the ball, we use the kinematics equation which relates the final velocity \( v \), initial velocity \( u \), acceleration \( a \), and displacement \( s \) (in this case, height \( h \)):</p> <p>\[ v^2 = u^2 + 2as \]</p> <p>At the maximum height, the final velocity \( v \) will be 0 m/s (since the ball stops rising before it starts to fall). The initial velocity \( u \) is 10 m/s (upward), and the acceleration \( a \) due to gravity is -10 m/s² (downward). Thus, we can plug these into our equation:</p> <p>\[ 0 = (10)^2 + 2(-10)h \]</p> <p>Which simplifies to:</p> <p>\[ 0 = 100 - 20h \]</p> <p>Now we solve for \( h \):</p> <p>\[ 20h = 100 \]</p> <p>\[ h = \frac{100}{20} \]</p> <p>\[ h = 5 \]</p> <p>Thus, the maximum height \( h \) reached by the ball is 5 meters.</p>
Calculating Launch Angle for Projectile Motion
The image contains a text in Italian that talks about a projectile being fired with an initial velocity, and it asks to calculate the launch angle such that the projectile can hit a target placed a certain distance away. Unfortunately, the full details of the problem are not completely visible, but I can provide a generic approach for such a problem. To solve for the launch angle (\(\theta\)) of a projectile given the initial velocity (\(v_0\)) and the distance to the target (\(R\)), we can use the following formula, assuming no air resistance and that the launch and target points are at the same vertical level: \[ R = \frac{v_0^2}{g} \sin(2 \theta) \] Where \( R \) is the range or distance to the target, \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity (which is approximately \( 9.81 m/s^2 \) on Earth), and \( \theta \) is the launch angle. Rearrange this equation to solve for \( \theta \): \[ \sin(2 \theta) = \frac{R g}{v_0^2} \] Take the inverse sine to find \( 2\theta \): \[ 2\theta = \sin^{-1}\left(\frac{R g}{v_0^2}\right) \] Then, divide by 2 to find the angle \( \theta \): \[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{R g}{v_0^2}\right) \] Now, input the given values (\( v_0 = 130 m/s \) and \( R = 1100 m \)), and solve for \( \theta \): \[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1100 \times 9.81}{(130)^2}\right) \] \[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{10791}{16900}\right) \] \[ \theta = \frac{1}{2} \sin^{-1}(0.6383) \] \[ \theta \approx \frac{1}{2} \times 39.86^\circ \] \[ \theta \approx 19.93^\circ \] The launch angle needed to hit the target 1100 meters away with an initial velocity of 130 m/s is approximately \( 19.93^\circ \). Keep in mind that due to the nature of the sin function there are two possible solutions for \( \theta \) in the range [0°, 90°], as sin(θ) = sin(180° - θ). Therefore, the other possible angle would be \( 90^\circ - 19.93^\circ = 70.07^\circ \). These two angles represent the low and high trajectories that both reach the target.
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