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 17 m, it has dropped 55 cm from its original height when it left the racquet. What was the initial speed, in m/s, of the tennis ball

Answer 1

Answer:

The value is $u_x = 56.1 \ m/s$

Explanation:

From the question we are told that

    The  horizontal distance traveled is  $h = 17 \ m$

    The decrease in height is  

Considering the vertical velocity

Generally from kinematic equation we have that

      $v_y ^2 = u_y ^2 + 2* g * s$

Here the initial vertical ( $u_y$)  velocity is  zero

=>   $v_y ^2 = 0 ^2 + 2 * 9.8 * 0.55$

=>  $v_y = 2.97 \ m$

Generally the time to drop to this height is mathematically represented as

      $t = \frac{v_y }{g }$

=>   $t = \frac{ 2.97 }{ 9.8 }$

=>   $t = 0.3031 \ s$

Generally given that time taken to drop to the obtained height is equal to the time taken to reach the horizontal distance , then

      $u_x = \frac{h}{t}$

Here $u_x$ is the initial horizontal velocity

     $u_x = \frac{ 17 }{ 0.3031 }$

=> $u_x = 56.1 \ m/s$