Answer to Question #95446 in Classical Mechanics for Jake

Question

3. A new game has been invented where the player has to combine a dive from a board with shooting of a basketball through a hoop (see figure below). As a student of PHY 101 you decide to take part in this game. During your practice sessions, you realize that while standing on the diving board 4.50 m above the water, you are able to score into the hoop which is a horizontal distance of 6.50 m away from the board and 1.00 m above ground level. (a) With what speed did you launch the basketball? (b) Now you step off the diving board and launch the ball in order to make a basket into a hoop that is a horizontal of 3.75 m, at what time after stepping off the board should you launch the ball? (c) What are the horizontal and vertical components of the ball’s velocity at the instant of launch?

Answer 1

a)

$$t=\sqrt{\frac{2(H-h)}{g}}=\sqrt{\frac{2(4.5-1)}{9.8}}=0.845\ s$$

$$v=\frac{d}{t}=\frac{6.5}{0.845}=7.69\frac{m}{s}$$

b)

$$t'=\frac{3.75}{7.69}=0.488\ s$$

$$T=t-t'=0.845-0.488=0.357\ s$$

c)

$$v_y=gT=(9.8)(0.357)=3.50\frac{m}{s}$$

$$v_x=v=7.69\frac{m}{s}$$