Question

Chapter 04, Problem 036 During a tennis match, a player serves the ball at 25.5 m/s, with the center of the ball leaving the racquet horizontally 2.55 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

Part A. When ball is served horizontally, then

Initial horizontal speed = V0x = V0*cos = 25.5*cos 0 deg = 25.5 m/s

Initial vertical speed = V0y = V0*sin = 25.5*sin 0 deg = 0 m/s

Now range in projectile motion is given by

R = V0x*t (Since there is no acceleration in horizontal direction, So horizontal velocity remains constant)

Given that net is 12.0 away from the player, So

t = R/V0x = 12.0/25.5 = 0.4706 sec

Now Using 2nd kinematic equation in vertical direction:

y = y0 + V0y*t + (1/2)*ay*t^2

ay = acceleration in vertical direction = -9.81 m/s^2

y0 = initial vertical position = 2.55 m

So,

y = 2.55 + 0*0.4706 + (1/2)*(-9.81)*0.4706^2

y = 1.46 m

Given that net is 0.900 m high, while at that horizontal distance ball is 1.46 m high, So ball will clear the net

Now distance between top of the net and ball will be:

d = y - 0.900 = 1.46 - 0.900

d = 0.56 m

Part B. Using same process

When ball is served 5.00 below the horizontal, then

Initial horizontal speed = V0x = V0*cos = 25.5*cos 5 deg = 25.4 m/s

Initial vertical speed = V0y = V0*sin = -25.5*sin 5 deg = -2.22 m/s (-ve sign because ball is launched below the horizontal)

Now range in projectile motion is given by

R = V0x*t (Since there is no acceleration in horizontal direction, So horizontal velocity remains constant)

Given that net is 12.0 away from the player, So

t = R/V0x = 12.0/25.4 = 0.4724 sec

Now Using 2nd kinematic equation in vertical direction:

y = y0 + V0y*t + (1/2)*ay*t^2

ay = acceleration in vertical direction = -9.81 m/s^2

y0 = initial vertical position = 2.55 m

So,

y = 2.55 + (-2.22)*0.4724 + (1/2)*(-9.81)*0.4724^2

y = 0.407 m

Given that net is 0.900 m high, while at that horizontal distance ball is 0.407 m high, So ball will not clear the net

Now distance between top of the net and ball will be:

d = y - 0.900 = 0.407 - 0.900

d = -0.493 m

(-ve sign means ball does not clear the net)

Let me know if you've any query.