Question #109408

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.

Expert's answer

Given:

v0=24.5m/s,v_0=24.5\:\rm m/s,

θ=60° ⁣,\theta=60\degree\!,

g=9.81m/s2.g=9.81\:\rm m/s^2.


(a) The time of flight

t=2v0sinθg=2×24.5sin60°9.81=4.33s.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=v02sin2θg=24.52sin120°9.81=53.0m.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

Hmax=v02sin2θ2g=24.52sin260°2×9.81=22.9m.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.
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