Question #258925

A bullet is fired from gun mounted at an angle of 30°. If the muzzle velocity is 400 m/s, calculate:

a. Vertical velocity component

b. Horizontal velocity component

c. Maximum height reach

d. Time of flight

e. Range


1
Expert's answer
2021-10-31T18:10:17-0400

Given:

v0=400m/sv_0=400\:\rm m/s

g=9.81m/s2g=9.81\:\rm m/s^2

θ=30\theta=30^{\circ}


(a) Vertical velocity component

vy=v0sinθ=400sin30=200m/sv_y=v_0\sin\theta=400*\sin 30^{\circ}=200\:\rm m/s

(b) Horizontal velocity component

vx=v0cosθ=400cos30=346m/sv_x=v_0\cos\theta=400*\cos 30^{\circ}=346\:\rm m/s

(c) Maximum height reach

hmax=v02sin2θ2g=4002sin23029.81=2039mh_{\max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{400^2\sin^230^{\circ}}{2*9.81}=2039\:\rm m

(d) Time of flight

t=2v0sinθg=2400sin309.81=40.8st=\frac{2v_0\sin\theta}{g}=\frac{2*400\sin30^{\circ}}{9.81}=40.8\:\rm s

(e) Range

hmax=v02sin2θg=4002sin609.81=12232mh_{\max}=\frac{v_0^2\sin2\theta}{g}=\frac{400^2\sin60^{\circ}}{9.81}=12232\:\rm m

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