Question

A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0° below horizontal. How far from the base of the building will the ball land?

Answer 1

Answer:

Explanation:

horizontal component of velocity of throw = 20 cos20 = 18.8 m /s

vertical downwards component = 20 sin20 = 6.84 m /s

time to displace by height 30 m = t  , initial velocity u = 6.84 m /s

h = ut + 1/2 gt²

30 = 6.84 t + .5 x 9.8 t²

4.9 t² + 6.84 t - 30 = 0

t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9

=  - 6.84 ±√( 46.78 + 588 ) / 9.8

=  - 6.84 ±√(634.78 ) / 9.8

= - 6.84 ±25.2 / 9.8

= 1.87 s

horizontal displacement in 1.87 s

= 18.8 x 1.87

= 35.15 m .