Answer to Question #109408 in Physics for Steven

Question

Suppose the body in Sample 16.2 is projected at 24.5 m/s, 60° above the horizontal. Find the a. time of flight, b. range, and c. Maximum height.

Answer 1

Given:

$v_0=24.5\:\rm m/s,$

$\theta=60\degree\!,$

$g=9.81\:\rm m/s^2.$

(a) The time of flight

$t=\frac{2v_0\sin\theta}{g}=\frac{2\times 24.5\sin 60\degree}{9.81}=4.33\:\rm s.$

(b) The range

$R=\frac{v_0^2\sin2\theta}{g}=\frac{24.5^2\sin 120\degree}{9.81}=53.0\:\rm m.$

(c) The maximum height

$H_{\rm max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{24.5^2\sin^2 60\degree}{2\times 9.81}=22.9\:\rm m.$