Question

During a tennis match, a player serves the ball at 25.6 m/s, with the center of the ball leaving the racquet horizontally 2.34 m above the court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number. Use g=9.81 m/s².

Answer 1

PROJECTILE

along horizontal
________________

initial velocity v0x = v

acceleration ax = 0

initial position = xo = 0

final position = x = 12 m

displacement = x - x0

from equation of motion

x - x0 = v0x*T + 0.5*ax*t^2

x - x0 = v*t

t = (x - x0)/(v) = 12/25.6 = 0.46875 s

along vertical
______________

initial velocity v0y = 0

acceleration ay = -g = -9.81 m/s^2

initial position y0 = 2.34 m

final position y = ?

from equation of motion

y-y0 = v0y*t + 0.5*ay*t^2

y-2.34 = 0 - (0.5*9.81*0.46875^2)

y = 1.262 m

distance between center of racquet and the top of net = 1.262-0.9 = 0.362 m

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PROJECTILE

along horizontal
________________

initial velocity v0x = v*costheta = 25.6*cos5

acceleration ax = 0

initial position = xo = 0

final position = x = 12 m

displacement = x - x0

from equation of motion

x - x0 = v0x*T + 0.5*ax*t^2

x - x0 = v*costheta*t

t = (x - x0)/(v) = 12/(25.6*cos5) = 0.471 s

along vertical
______________

initial velocity v0y = -v*sintheta

acceleration ay = -g = -9.81 m/s^2

initial position y0 = 2.34 m

final position y = ?

from equation of motion

y-y0 = v0y*t + 0.5*ay*t^2

y-2.34 = -(25.6*sin5*0.471) - (0.5*9.81*0.471^2)

y = 0.201 m

distance between center of racquet and the top of net = 0.201-0.9 = -0.699 m