A particle is projected with a velocity of 16.0 ft/s at an angle of 20.0 degree

Question

A particle is projected with a velocity of 16.0 ft/s at an angle of 20.0 degree below the horizontal, The particle passes through a small hole in a vertical fence at a height Y_p above the ground and at a horizontal distance X_p to the right of the firing point. As its passing through the hole it is traveling at 53.13 degree below the horizontal. After passing through the hole the particle lands at a point 120 ft below the launching point. (a) What are the X_p and Y_p coordinates of the small hole? (b) Determine the horizontal range X_H

Explanation / Answer

In horizontal,

V0x = 16 cos20 = 15.03 ft/s

there is no acceleration in horizontal hence it will move with constant velocity in horizonatal.

In vertical,

v0y = - 16 sin20 = - 5.47 ft/s

vy = v0y + ay t = - 5.47 - 32.17t

{ 32.17 ft/s^2 is acc. due to gravity}

at point P, tan53.13 = (5.47 + 32.17t) / 15.03

20.04 = 5.47 + 32.174t

t = 0.453 sec

(A) Xp = (v0x)(t) = 15.03 x 0.453 = 6.81 ft

Yp = 120 - (5.47 x 0.453 + 32.17 0.453^2 /2) = 114.2 ft

(B) to reach the ground,

vertical displacement = -120 ft

-120 = - 5.47t - 32.174 t^2 /2

16.1t^2 + 5.47t - 120 = 0

t = 2.57 sec

XH = v0x t = 15.03 x 2.57 = 38.6 ft

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