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. (b) determine the maximum height. 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 s velocity m/s (d) determine the time needed for the ball to reach the ground. your response is within 10% of the correct value. this may be due to round - off error, or you could have a mistake in your calculation. carry out all intermediate results to at least four - digit accuracy to minimize roundoff error. s (e) determine the velocity and position of the ball at t = 5.70 s. velocity m/s position m the response you submitted has the wrong sign. m

Explanation:

Step1: Recall kinematic - equation for maximum height

At maximum height, the final velocity $v = 0$. The kinematic equation $v=v_0 - gt$ is used to find the time to reach maximum - height, where $v_0 = 24.6$ m/s and $g = 9.8$ m/s².

Step2: Calculate the time to reach maximum height

$t=\frac{v_0 - v}{g}=\frac{24.6-0}{9.8}\approx2.51$ s.

Step3: Recall kinematic - equation for displacement

The kinematic equation for displacement $y - y_0=v_0t-\frac{1}{2}gt^{2}$ is used to find the maximum height. We know $y_0 = 53.3$ m, $v_0 = 24.6$ m/s, and $t = 2.51$ s.

Step4: Calculate the maximum height

$y=y_0+v_0t-\frac{1}{2}gt^{2}=53.3 + 24.6\times2.51-\frac{1}{2}\times9.8\times(2.51)^{2}$
$y = 53.3+61.746 - 30.87009\approx84.176$ m.

Step5: Analyze the motion back to the starting height

The time to go up is the same as the time to come back down to the same height. So the time to return to the starting height is $t = 2\times2.51 = 5.02$ s. Using $v = v_0 - gt$ with $t = 5.02$ s and $v_0 = 24.6$ m/s, $v=24.6-9.8\times5.02=24.6 - 49.196=-24.6$ m/s.

Step6: Recall kinematic - equation for displacement to the ground

The kinematic equation $y - y_0=v_0t-\frac{1}{2}gt^{2}$, where $y = 0$, $y_0 = 53.3$ m, and $v_0 = 24.6$ m/s. So, $0 = 53.3+24.6t-4.9t^{2}$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$ (here $a=-4.9$, $b = 24.6$, $c = 53.3$).
$t=\frac{-24.6\pm\sqrt{(24.6)^{2}-4\times(-4.9)\times53.3}}{2\times(-4.9)}=\frac{-24.6\pm\sqrt{605.16 + 1044.68}}{-9.8}=\frac{-24.6\pm\sqrt{1649.84}}{-9.8}=\frac{-24.6\pm40.62}{-9.8}$. We take the positive root $t=\frac{-24.6 + 40.62}{-9.8}\approx6.65$ s.

Step7: Recall kinematic - equations for velocity and position at a given time

For velocity, $v = v_0 - gt$, with $v_0 = 24.6$ m/s, $g = 9.8$ m/s², and $t = 5.70$ s. $v=24.6-9.8\times5.70=24.6 - 55.86=-31.26$ m/s.
For position, $y=y_0+v_0t-\frac{1}{2}gt^{2}$, with $y_0 = 53.3$ m, $v_0 = 24.6$ m/s, $t = 5.70$ s, and $g = 9.8$ m/s².
$y = 53.3+24.6\times5.70-\frac{1}{2}\times9.8\times(5.70)^{2}=53.3 + 140.22-158.769 = 34.751$ m.

Answer:

(b) $84.176$ m
(d) $6.65$ s
(e) Position: $34.751$ m