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Question

http://ef.engr.utk.edu/c/sys/13/assess/paramimagev2.php?n=/www/c/ef151-2012-01/img/assess/tennis-trajectory.png&p_AB=36,406,223,0&p_AH=6.7,641,115,0&p_BH=3.2,288,150,0

A) Determine the speed of the tennis ball at A. (ft/sec)
B) Determine the distance s where the ball hits the ground. (ft)
C) Determine the speed of the ball the instant before it hits the ground. (ft/sec)
D) Determine the angle ? that represents the angle between the ball's velocity vector and the ground. (°, acute, positive - see picture)

Explanation / Answer

A)

Vertical distance travelled just before reaching the net = 6.7-3.2 = 3.5 ft

we know only acceleration due to gravity isin the verica direction

so g = 32.15 ft/s^2

1/2*g*t^2 = 3.5 ft

t = (3.5*2/(32.15)) = 0.467 s

horizontal distance travelled in this time = V*t = 36 ft

V*0.467 = 36

V = 36/0.467 = 77.15 ft/sec

B)

vertical distance travelled when ball hit the ground = 1/2*g*t^2 = 6.7 ft

t = (2*6.7/(32.15)) = 0.6456 sec

horizontal distance travelled = V*t = 77.15*0.6456 = 49.8 ft

s = 49.8-36 = 13.8 ft

C)

horizontal speed at the time of hitting the ground = 77.15 ft/sec

Vertical speed = g*t = 32.15*0.6456 = 20.756 ft/sec

speed = (77.15^2+20.756^2) = 79.89 ft/sec

D)

Angle = tan^-1(20.756/77.15) = 15.05 degrees

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