Answer to Question #143165 in Physics for Erica

Question

A ball thrown horizontally from the top of a building 55m high strikes the ground at a point 35m for om the building. Find the (A) time to reach the ground (B) the initial speed of the ball, and (C) the velocity with the ball will strike the ground.

Answer 1

(A) We can find the time to reach the ground from the kinematic equation:

$$y=\dfrac{1}{2}gt^2,$$

here, $y=55\ m$ is the vertical displacement of the ball (or the height of the building), $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity, $t$ is the time to reach the ground.

Then, from this equation we can calculate $t$:

$$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot 55\ m}{9.8\ \dfrac{m}{s^2}}}=3.35\ s.$$

(B) We can find initial horizontal velocity of the ball from the kinematic equation:

$$x=v_{0x}t,$$

here, $x=35\ m$ is the horizontal displacement of the ball (or the horizontal range).

Than, we get:

$$v_{0x}=\dfrac{x}{t}=\dfrac{35\ m}{3.35\ s}=10.45\ \dfrac{m}{s}.$$

(C) Let's firs find final vertical velocity from the kinematic equation:

$$v_f=v_i-gt=0-9.8\ \dfrac{m}{s^2}\cdot 3.35\ s=-32.83\ \dfrac{m}{s}.$$

The sign minus means that the final vertical velocity directed downward.

Finally, from the Pythagorean theorem we can find the velocity with which the ball will strike the ground:

$$v=\sqrt{v_{fx}^2+v_{fy}^2}=\sqrt{(10.45\ \dfrac{m}{s})^2+(-32.83\ \dfrac{m}{s})^2}=34.45\ \dfrac{m}{s}.$$

Answer:

(A) $t=3.35\ s.$

(B) $v_{0x}=10.45\ \dfrac{m}{s}.$

(C) $v=34.45\ \dfrac{m}{s}.$