Question

use the worked example above to help you solve this problem. a ball is thrown from the top of a building with an initial velocity of 24.6 m/s straight upward, at an initial height of 53.3 m above the ground. the ball just misses the edge of the roof on its way down, as shown in the figure. (a) determine the time needed for the ball to reach its maximum height. enter a number. (b) determine the maximum height. (c) determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. time s velocity m/s (d) determine the time needed for the ball to reach the ground. (e) determine the velocity and position of the ball at t = 5.70 s. velocity m/s position m

Explanation:

Step1: Find time to max - height

Use $v = v_0+at$. At max - height, $v = 0$. $a=-g=- 9.8\ m/s^2$, $v_0 = 24.6\ m/s$. So $t=\frac{v - v_0}{a}=\frac{0 - 24.6}{-9.8}$.

Step2: Calculate max - height

First, find the height increase $\Delta h$ using $v^{2}-v_{0}^{2}=2a\Delta h$. Then $h_{max}=h_0+\Delta h$, where $h_0 = 53.3\ m$.

Step3: Time to return to initial height

The motion is symmetric. The time to go up is the same as the time to come down to the initial height. So the time is $2t$ (where $t$ is from step 1). The velocity at that instant is $v=-v_0$.

Step4: Time to reach ground

Use $y - y_0=v_0t+\frac{1}{2}at^{2}$, where $y = 0$, $y_0 = 53.3\ m$, $v_0 = 24.6\ m/s$, $a=-9.8\ m/s^{2}$. Solve the quadratic equation $-53.3=24.6t-4.9t^{2}$ for $t$.

Step5: Velocity and position at $t = 5.70\ s$

Use $v = v_0+at$ to find velocity and $y - y_0=v_0t+\frac{1}{2}at^{2}$ to find position, with $y_0 = 53.3\ m$, $v_0 = 24.6\ m/s$, $a=-9.8\ m/s^{2}$ and $t = 5.70\ s$.

Answer:

(a) $\frac{0 - 24.6}{-9.8}\approx2.51\ s$
(b) First, $\Delta h=\frac{0 - 24.6^{2}}{2\times(-9.8)}\approx30.9\ m$, $h_{max}=53.3 + 30.9=84.2\ m$
(c) Time: $2\times2.51 = 5.02\ s$, Velocity: $- 24.6\ m/s$
(d) Solving $-53.3=24.6t-4.9t^{2}$, using the quadratic formula $t=\frac{-24.6\pm\sqrt{24.6^{2}-4\times(-4.9)\times(-53.3)}}{2\times(-4.9)}$. We get $t\approx6.91\ s$ (we take the positive root).
(e) Velocity: $v=24.6-9.8\times5.70=24.6 - 55.86=-31.26\ m/s$. Position: $y-53.3=24.6\times5.70-4.9\times5.70^{2}$, $y=53.3+24.6\times5.70-4.9\times5.70^{2}=53.3 + 140.22-158.343 = 35.177\ m$