Question

You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. (Figure 1) Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend. What minimum speed will allow the ball to clear the roof? At what angle should you toss the ball?

Answer 1

Part A:

Maximum height to be reached by ball = 3+3-1=5m

Horizontal distance to be reached by ball = 18m

Let the ball be thrown at speed v and at an angle $\theta$

5=(v²sin² $\theta$ )/2g .... (i)

18=v²sin2 $\theta$ /g...(ii)

dividing(ii) by (i) we get

18/5 = 2(sin2 $\theta$ )/sin² $\theta$

1.8=(2sin $\theta$ cos $\theta$ )/sin² $\theta$

0.9=1/tan $\theta$

tan $\theta$ = 1/0.9

$\theta$ = 48^o

5=(v²sin² $\theta$ )/2g

Or, 5=(v²sin²48)/2g

v = 13.32 m/s

Part B :

$\theta$ = 48^o