# When a tennis player hits a ball its velocity is 10 m/s at an angle of 30 degrees above the horizontal. The ball just clears a net 7 m away that is 1 m high. At what height was the ball hit.

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The tennis ball was hit by the player resulting in a velocity of 10 m/s at an angle 30 degrees above the horizontal. The ball travels to clear a net 7 m away that has a height of 1 m. The height at which the ball was hit has to be determined. Let the height be H.

When the ball is hit the vertical component of its velocity is 10*sin 30 and the horizontal component is 10*cos 30. The time taken by the ball to reach the net is t = 7/10*cos 30. In this duration the ball rises from a height H and drops to a height of 1 m. This gives: H - 1 = 5*(7/10*cos 30) - 0.5*9.8*(7/10*cos 30)^2

=> H = 1 + 5*(7/10*cos 30) - 0.5*9.8*(7/10*cos 30)^2

=> H = 1.84 m

The ball had to be struck at a height of 1.84 m for it to just cross the net.

**Sources:**

for this question we can use the following equation,

S = u.t + (1/2).a.(t^2) .... (1)

in the x direction the velocity component is given by, 10.cos30

= 10x0.867

= 8.67 m/s

In the x direction the distance to the net is 7m.

Lets calculate the time taken to reach the net by applying the equation (1) in the x direction

S = u.t + (1/2).a.(t^2) .... (1)

=> 7 = 8.67 t + 0 (acceleration in the x direction is zero , a =0)

=> t = 0.807 s

Now lets calculate the verticle distance the ball has travelled at that time by applying the equation in the y direction

velosity component in the y direction is = 10. sin 30

= 10 x (0.5)

= 5 m/s

if the height of the projection is h from the ground, the verticle distance it has travelled is given by, (h-1) as at this point it is 1 m above the ground

S = u.t + (1/2).a.(t^2) .... (1)

=> h - 1 = (5 m/s).(0.807 s) + (0.5)(-9.81 m/s.s)(0.807^2 s.s)

here the acceleration due to gravity is a negative value in the positive y direction

= > h =1 + 4.035 - 3.194

=> ** h= 1.841 m**