Dynamics Question, Im stuck A table tennis player hits a ball horizontally(strai
ID: 1857931 • Letter: D
Question
Dynamics Question, Im stuck
A table tennis player hits a ball horizontally(straight toward the net) at a speed of 6.8 m/s. The player strikes the ball 37.5 cm above the table surface while standing 2.37 m from the net. The ball bounces once, and just clears the 15.25-cm high net. Determine the time required for the ball to bounce on the table t12 the horizontal distance travelled x12 the velocity if ball when it bounces(vx, 2 and vy, 2) the horizontal distance from the first bounce to the net x34 the time required for the ball to travel the remaining distance to the net t34 the velocity of the ball after impact(vx, 3 and vy, 3) in terms of the coefficient of restitution e the coefficient of restitution of the ball if the ball just clears the net (hnet = 15.25cm) Neglect the aerodynamics and physical size of the ball.Explanation / Answer
a) y = y0 + v0y t + 1/2 at^2
0 = .375-0.5*9.81*t^2
t=0.2765 s
b) x = vt = 6.8*0.2765=1.8802 m
c) vx,2 = 6.8
vy,2 = at = -9.81*0.2765=-2.712
d) x34 = 2.37-1.8802=0.4898 m
e) vx doesnt change upon bounce since parallel to surface
t = x34/v = 0.4898/6.8=0.072 s
f) vx3 = v32 = 6.8
e*sqrt(vx,2^2 + vy,2^2) = sqrt(vy,3^2 + vx,3^2)
e^2*sqrt(6.8^2+2.712^2) = sqrt( vy^2 + 6.8^2)
g) y direction
y = y0 + v0y t + 1/2 at^2
.1525 = v*0.072-0.5*9.81*0.072^2
v=2.4712 m/s
e^2*sqrt(6.8^2+2.712^2) = sqrt( 2.4712^2 + 6.8^2)
e = 0.994126
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