A team of engineering students designs a medium-size catapult which launches 8-l

Question

A team of engineering students designs a medium-size catapult which launches 8-lb steel spheres. The launch speed is upsilon_0 = 80 ft/sec, he launch angle is theta = 35 degree above the horizontal, and the launch position is 6 ft above ground level. The students use an athletic field with in adjoining slope topped by an 8-ft fence as shown. Determine: (a) the x-y coordinates of the point of first impact (b) the time duration t_f of the flight (C) the maximum height h above the horizontal field attained by the ball (d) the velocity (expressed as a vector) with which the projectile strikes the ground Repent part (a) for a launch speed of upsilon_0 = 75 ft/sec.

Explanation / Answer

max height reached by projectile= 80^2 sin ^2 ( 35)/ 32= 65.79 ft apprx

time of half flight = 2( 80) sin 35/32= 2.8678 sec apprx

time taken by projectile to cover a height of 65.79 - 22 =43.79 ft

43.79 ft = 0.5 (32) t^2

t= 1.654 sec

horizontal distance covered during ( 2.8678 + 1.654 )= 296.29 ft

( x, y coordinate of imapct ) = ( 296.29, 22)

b) time duration of flight = 4.5218 sec apprx

c) max height = 65.79 ft

d) velocity after falling 43.79 feet = sqroot ( 2x 32 x 43.79) = 52.94 ft /sec

V at srike= sqroot ( Vx^2 + Vy^2) = 84.244 ft/sec

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