Question

A golf ball with an initial angle of 32° lands exactly 224 m down the range on a level course.

(a) Neglecting air friction, what initial speed would achieve this result? [m/s]
(b) Using the speed determined in item (a), find the maximum height reached by the ball. [m]

Answer 1

Step-by-step explanation:

Given

launch angle
`=32^(\circ)`

ball lands exactly 224 m

Range of projectile =224 m

`Range =(u^2sin2\theta )/(g)`

`224=(u^2sin64)/(9.8)`

`u^2sin64=224* 9.8=2195.2`

`u^2=2442.383`

u=49.42 m/s

(b)Maximum height reached

`H_(max)=(u^2sin^2\theta )/(2g)`

`H_(max)=(49.42^2* sin^(2)32)/(2\cdot 9.8)`

`H_(max)=34.99 m\approx 35 m`