Question

The third floor window is 50 meters away horizontally, and 12 meters above the ground. If you launch a tennis ball at the initial speed of 62.5 m/sec (140mph) at what angle(s) must you launch the ball so that it will enter the third flood window? Please show work

Answer 1

Using the equation of trajectory

y = r tan 0-3 2(u cos 62

where y = 12 m, x = 50 m, and v = 62.5 m/s

12 m = 50 mtan - 9.8 m/s2 x (50 m2 2(62.5 m/s x cos 6)?

12 m = 50 m tan - (3.14 sec )

12 m = 50 m tan - 3.14(1+tan? )

3.14 tane - 50 tan 6 + 15.14 = 0

Solving this quadratic equation we get

tan 0 = 15.6 :0.31

A = 17.2° Or 86.34