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. 2.51 s (b) determine the maximum height. 84.176 your response differs from the correct answer by more than 100%. m (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 5.02 s velocity -24.6 m/s (d) determine the time needed for the ball to reach the ground. 6.65 s (e) determine the velocity and position of the ball at t = 5.70 s. velocity -31.26 m/s position 34.751 the response you submitted has the wrong sign. m

Explanation:

Step1: Recall the kinematic equation for vertical - motion

The kinematic equation for vertical displacement is $y = y_0+v_0t-\frac{1}{2}gt^2$, where $y_0$ is the initial height, $v_0$ is the initial velocity, $t$ is the time, and $g = 9.8\ m/s^2$. Here, $y_0=53.3\ m$, $v_0 = 24.6\ m/s$, and we want to find $y$ at $t = 5.70\ s$.

Step2: Substitute the values into the equation

$y=53.3+24.6\times5.70-\frac{1}{2}\times9.8\times(5.70)^2$.
First, calculate the individual terms:
$24.6\times5.70 = 140.22$.
$\frac{1}{2}\times9.8\times(5.70)^2=4.9\times32.49 = 159.201$.
Then, $y=53.3 + 140.22-159.201$.
$y=34.319\ m$.

Answer:

$34.319$