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

Explanation:

Given

launch angle$=32^{\circ}$

ball lands exactly 224 m

Range of projectile =224 m

$Range =\frac{u^2sin2\theta }{g}$

$224=\frac{u^2sin64}{9.8}$

$u^2sin64=224\times 9.8=2195.2$

$u^2=2442.383$

u=49.42 m/s

(b)Maximum height reached

$H_{max}=\frac{u^2sin^2\theta }{2g}$

$H_{max}=\frac{49.42^2\times sin^{2}32}{2\cdot 9.8}$

$H_{max}=34.99 m\approx 35 m$