Answer to Question #169146 in Physics for Laureeeee

Question

An entertainer juggles balls while doing other activities. In one act, she throws a ball

vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at a

constant speed of 2.50 m/s, returning just in time to catch the falling ball.

(a) With what

minimum initial speed must she throw the ball upward to accomplish this feat?

(b) How high

above its initial position is the ball just as she reaches the table?

Answer 1

a)

$$t=\frac{d}{v}=\frac{2(5.5)}{2.5}=4.4\ s$$

$$0=v_0t-0.5gt^2\\v_0=0.5(9.8)(4.4)=21.6\frac{m}{s}$$

b)

$$t=\frac{(5.5)}{2.5}=2.2\ s$$

$$y=v_0t-0.5gt^2\\y=(21.6)(2.2)-0.5(9.8)(2.2)^2=23.8\ m$$