Hitting a Tennis Ball

Question

A tennis player hits a ball 2.0m above the ground. The ball leaves his racquet with a speed of 25.0 m/s at an angle 5.5 degrees above the horizontal. The horizontal distance to the net is 7.0m , and the net is 1.0m high.

1.) Does the ball clear the net?

2.) By how much?

Can someone please show me how to set this up!!

Answer 1

Ok, so you know there is no acceleration in the x-direction.  So what you need to do is find out when the ball is at the net to see if it is over or under 1.0 m.   So, the velocity in the x direction is 25*cos5.5.  Now you can find the time at which it reaches the net.  You know the net is 7.0 m away.  

So.. V = d/t   t = 0.281 s

Now you need to know where the vertical position of the ball is at this time.  

Xf = Xi + Vt + .5*a*t^2

xf == final position

xi == initial position

V == initial velocity

a == vertical acceleration

t == time

Xf = ?

Xi = 2m

V = 25*sin5.5

t = 0.281

a = -9.81 m/s^2

xf = 2.286 m

The ball clears the net by 1.286 m

Answer 2

You need the clearance at X = 7 m from the player.  vx = V cos(theta); where V = 25 mps and theta = 5.5.  vx = constant if we discount air drag (which we have to do since you gave no drag data).  Thus the time to get over the 7 m to the net is tx = X/vx = X/(V cos(theta)) and you can solve for how long the ball needs to be above Y > 1.0 meter.

Find max height above the point where the ball was struck.  h = vy^2/2g = (V sin(theta))^2/2g; where g = 9.81 m/sec^2.  theta = 5.5 degrees.  Find tup = sqrt(2h/g), the time to reach max height h above the launch point.  

tup is also the same time to reach H = h + 2, which is the max height above ground when the strike point is 2 m high.  If tup >= tx, we know the ball is at least Y > 2 meter;; so the ball will clear the net.  If tup < tx, we need to find the time tdown, where that's the time to fall from H to Y = 1 meter the net height.

tdown = sqrt(2(H - 1)/g) = sqrt(2(h + 2 - 1)/g) = sqrt(2(h + 1)/g), you found h earlier; so you can find tdown.  Then if tup + tdown > tx, the ball will clear the net...othewise game, set, and match to your opponent.

Net clearance H - S(tdown) - 1 = clearance; where H - S(tdown) is the distance above the ground after the ball falls for tdown seconds and the - 1 is the height of the net so the clearance > 0 is above the net.  H = h + 2 and S(tdown) = 1/2 g(tdown)^2 you found h and tdown earlier.  You can do the math.