Question

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a horizontal distance of 13 m, it has dropped 57 cm from its original height when it left the racquet. What was the initial speed of the tennis ball? (Neglect air resistance.)

Answer 1

Answer:

$u_x=38.13\ m/s$

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

$S=u_yt+\dfrac{1}{2}gt^2$

Here $u_y=0$

S= 57 cm

$0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=0.34 s

Now in the horizontal direction

$x=u_xt$

Here x=13 m

t= 0.34 s

So

$13=u_x\times 0.34$

$u_x=38.13\ m/s$

So the initial speed of ball is 38.13 m/s.