Question

1. a steel ball is thrown from the ledge of a tower so that it has an initial velocity of magnitude 15.0 m/s. the ledge is 15.0 m above the ground. air resistance is negligible. (a) what are the total flight time and the speed of impact at the ground if the initial velocity is upward? (b) what are these two quantities if the initial velocity is downward? (c) based on your answers to (a) and (b), write a concluding statement.

Explanation:

Step1: Set up the kinematic - equation for vertical motion

The kinematic equation for vertical displacement $y - y_0=v_0t-\frac{1}{2}gt^2$, where $y - y_0=- 15.0\ m$ (taking down - ward as negative, $y_0 = 15.0\ m$ and $y = 0$), $v_0$ is the initial vertical velocity, $g = 9.8\ m/s^2$, and $t$ is the time of flight. Also, the kinematic equation for final velocity $v = v_0−gt$.

Step2: Solve for time when initial velocity is upward ($v_0 = 15.0\ m/s$)

Substitute into $y - y_0=v_0t-\frac{1}{2}gt^2$:
$-15=15t - 4.9t^2$. Rearranging gives $4.9t^2-15t - 15 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ for the quadratic equation $ax^2+bx + c = 0$. Here, $a = 4.9$, $b=-15$, and $c=-15$.
$t=\frac{15\pm\sqrt{(-15)^2-4\times4.9\times(-15)}}{2\times4.9}=\frac{15\pm\sqrt{225 + 294}}{9.8}=\frac{15\pm\sqrt{519}}{9.8}=\frac{15\pm22.8}{9.8}$. We take the positive root $t=\frac{15 + 22.8}{9.8}=\frac{37.8}{9.8}\approx3.86\ s$.
Then, find the final velocity $v = v_0−gt=15-9.8\times3.86=15 - 37.83=-22.83\ m/s$. The speed of impact is $|v|\approx22.8\ m/s$.

Step3: Solve for time when initial velocity is downward ($v_0=-15.0\ m/s$)

Substitute into $y - y_0=v_0t-\frac{1}{2}gt^2$:
$-15=-15t-4.9t^2$. Rearranging gives $4.9t^2 + 15t-15 = 0$. Using the quadratic formula $t=\frac{-15\pm\sqrt{15^2-4\times4.9\times(-15)}}{2\times4.9}=\frac{-15\pm\sqrt{225 + 294}}{9.8}=\frac{-15\pm\sqrt{519}}{9.8}=\frac{-15\pm22.8}{9.8}$. We take the positive root $t=\frac{-15 + 22.8}{9.8}=\frac{7.8}{9.8}\approx0.80\ s$.
Then, find the final velocity $v = v_0−gt=-15-9.8\times0.80=-15 - 7.84=-22.84\ m/s$. The speed of impact is $|v|\approx22.8\ m/s$.

Step4: Write the concluding statement

The speed of impact at the ground is the same in both cases (a) and (b) because the mechanical - energy is conserved. The time of flight is different, with a longer time when the ball is thrown upward and a shorter time when the ball is thrown downward.

Answer:

(a) Flight time: $t\approx3.86\ s$, Speed of impact: $v\approx22.8\ m/s$
(b) Flight time: $t\approx0.80\ s$, Speed of impact: $v\approx22.8\ m/s$
(c) The speed of impact at the ground is the same whether the initial velocity is upward or downward due to energy conservation, but the flight - time is different.