Question

1.   A volleyball player serves a ball with his height of 1.5-meter height and strikes the floor at point three meter horizontally.

a.   Time of flight of the ball

b.   Its horizontal velocity

Answer 1

Explanations & Calculations

$\qquad\qquad \begin{aligned} \to s&=ut\\ \small R&=\small v_xt\to v_x =\frac{R}{t}\cdots(1)\\ \\ \uparrow s&=ut+\frac{1}{2}at^2\\ \small -h&=\small v_yt+\frac{1}{2}(-g)t^2\cdots(2)\\ \\ \small \tan\theta&=\small \frac{v_y}{v_x}\cdots(3)\\ \end{aligned}$

• By eliminating $\small v_x\, \&\,v_y$ from those equations,

$\qquad\qquad \begin{aligned} \small v_y&=\small \frac{R}{t}\tan\theta\\ \\ \small -h&=\small R\tan\theta-\frac{1}{2}gt^2\\ \small &=\small \sqrt{\frac{2(R\tan\theta+h)}{g}} \end{aligned}$

• If the delivery was done horizontally, then the velocity components become like this and the final answer follows accordingly.

$\qquad\qquad \begin{aligned} \small \theta&=\small 0\\ \small \tan\theta &=\small 0\\ \small v_y&=\small 0\\ \\ \small t&=\small \sqrt{\frac{2h}{g}}\\ &=\small \sqrt{\frac{2\times1.5}{9.8}}\\ &=\small 0.6\,s\\ \\ \small v_x&=\small \frac{3.0m}{0.6s}\\ &=\small 5ms^{-1} \end{aligned}$