Question #349438

A body is projected downward at an angle of 30° with the horizontal from the

top of a building 170 m high.

Its initial speed is 40 m/s. (a) How long will it

take before striking the ground? (b) How far from the foot of the building will it

strike? (c) At what angle with the horizontal will it strike?





1
Expert's answer
2022-06-09T09:59:19-0400
x(t)=v0cosθt=40cos30t=34.6t  mx(t)=v_0\cos\theta*t\\ =40*\cos30^\circ*t=34.6t\;\rm m

y(t)=y0v0sinθtgt2/2=17040sin30t9.8t2/2=17020t4.9t2  my(t)=y_0-v_0\sin\theta*t-g*t^2/2\\ =170-40*\sin30^\circ*t-9.8*t^2/2\\ =170-20t-4.9t^2\;\rm m

(a) at the ground

y(t)=17020t4.9t2=0y(t)=170-20t-4.9t^2=0

So

t=4.2st=4.2\:\rm s

(b)

x(t)=34.64.2=145mx(t)=34.6*4.2=145\:\rm m

(c)

vx=v0cosθ=34.6m/svy=v0sinθ+gt=20+9.84.2=61m/sv_x=v_0\cos\theta=34.6\:{\rm m/s}\\ v_y=v_0\sin\theta+gt=20+9.8*4.2=61\:\rm m/s

θ=tan1vyvx=tan16134.6=60\theta=\tan^{-1}\frac{v_y}{v_x}=\tan^{-1}\frac{61}{34.6}=60^\circ


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