Velocity Q - Lack of Information?

Question

Hey, I'm really struggling with a problem and help on it would be MUCH appreciated. I feel like there isn't enough info but apparently, there is.

Q. Tennis player A is standing at the net so tennis player B (@ baseline) lobs the ball from near the ground aiming for the baseline on A's side 24m away. B hits the ball at an angle 60 degrees above horizontal.

1/ How fast was the ball going just after B hits the lob?

A does a backhand and returns the ball from the baseline just as it is about to contact the ground. The tennis ball skims the 0.9m high net at the top of its flight, landing 24m away.

2/ How fast is the ball travelling when it leaves A's racquet?

Thanks..

Answer 1

It seems the problem has enough information, however I agree that  delivery perhaps could be a bit clearer.

The ball starts at ground level at 60 degrees  above horizontal and just about lands 24 meters away.

1. The ball goes up and comes down. The 'down' time  t2 or free fall of the ball is equal to  the 'up' time t1.

t2= Vv/g

Vv- initial vertical velocity

g - acceleration due to gravity

Let t1=t2=t we have

S= 2tVcos(60)

t=Vsin(60) / g

now

S= 2(V sin(60) /g)Vcos(60) and finally [using sin(2A) =2 sin(A) cos(A)]

V=sqrt( Sg/( sin(120))

V= sqrt(24 x 9.81) / sin(120))=

V=16.5 m/s

2. Similarly you can do the second part.

At S = 12 m it is at a height 0.9 m and then continues for 24 m. We could  use earlier equation  V=sqrt( Sg/( sin(A)) however angle A is unknown. However we know that the time it took to clear the net (going up or vertical component) equal to the time approaching the net ( horizontal component)

Vv= Vsin(A);  Vv= gt; t=Vv/g;

Vh= Vcos(A) Vh= s/t also

The net's height is h and

h=0.5gt^2 then

t= sqrt(2h/g) follows that

Vv= sqrt(2hg)=sqrt(2 x 0.9 x 9.81)=4.2m/s

Vh= s/sqrt(2h/g) = s sqrt(g/2h)= 12sqrt(9.81/(2x0.9))= 28.0m/s

A= actan(Vv/Vh)=arctan(4.2/28)= 8.53 deg

V= sqrt(Vv^2 +Vh^2)=sqrt(801)=28.3 m/s

Answer 2

1)

Let u = velocity with which B had hit the ball.

Using the range formula, R = u^2 sin 2θ / g

=> u^2 = (gR) / sin 2θ

=> u^2 = (9.8*24) / sin (120°)

=> u = √[(9.8*24) / sin (120°)]

=> u = 16.5 m/s

2)

R = u^2 sin 2θ / g and H = u^2 sin^2 θ/2g

=> R/H = 4 cot θ

=> tan θ = 4H/R = 4*0.9/24 = 0.15

=> θ = 8.53°

Again using the range formula,

u^2 = (gR) / sin 2θ = (9.8 x 24) / sin (17.06°)

=> u = √[(9.8 x 24) / sin (17.06°)]

= 28.3 m/s.