Question

A tennis ball served horizontally from a height of $1.98{\text{ m}}$ just clears a net $0.9{\text{ m}}$ high and $12.2{\text{ m}}$ from the server. Where will the ball strike the ground? $g = 9.8{\text{ m}}{{\text{s}}^{ - 1}}$
A. $12.2{\text{m}}$ from the server
B. $16.6{\text{m}}$ from the server
C. $22.3{\text{m}}$ from the server
D. $24.5{\text{m}}$ from the server

Answer 1

We are given a tennis ball served horizontally and the height of the net and the distance between the server and the net and we need to find the distance from the server where the ball will strike the ground. Here we will use the equation of motion for horizontal projectile motion and then will find the distance between the server and the point where the ball will strike the ground.

Complete step by step answer:
Since the ball is projected horizontally, suppose horizontal velocity be u and since in downward direction the ball will fall freely with zero initial velocity under gravity. Let ${t_{total}}$ be the time ball takes from point of releasing to the point it strike on ground then, In vertical direction total distance covered by ball is $1.98{\text{ m}}$ then using formula $S = ut + \dfrac{1}{2}g{t_{total}}^2$ and since in vertical direction $u = 0$
so,
$1.98 = 0 + \dfrac{1}{2}(9.8){t_{total}}^2$
$\Rightarrow 1.98 = (4.9){t_{total}}^2$
$\Rightarrow {t_{total}} = \sqrt {\dfrac{{198}}{{490}}} \sec \to (i)$

Now, as the ball just passes the $0.9m$ high net which is at $12.2m$ distance from server horizontally, the ball covers a vertical distance of $1.98 - 0.9 = 1.08m$ at that point, and suppose t be the time ball takes to reach at this point then for vertical distance of $1.08\,m$ we have,
$S = ut + \dfrac{1}{2}g{t_{total}}^2$
$\Rightarrow 1.08 = 0 + \dfrac{1}{2}(9.8){t^2}$
$\Rightarrow \dfrac{{1.08}}{{4.9}} = {t^2}$
$\Rightarrow t = \sqrt {\dfrac{{108}}{{490}}\sec }$
and for this time the ball simultaneously cover a horizontal distance of $12.2\,m$ with an initial velocity of $u$ then we have,
$S = ut$ put value of $t = \sqrt {\dfrac{{108}}{{490}}\,\sec }$
$\Rightarrow 12.2 = u\sqrt {\dfrac{{108}}{{490}}}$

We get,
$u = 12.2\sqrt {\dfrac{{490}}{{108}}} m{s^{ - 1}}$
now with this initial velocity, let us suppose ball strike the ground at a distance of S from the server then total time to reach the ground is already calculated by equation (i) as ${t_{total}} = \sqrt {\dfrac{{198}}{{490}}} \,\sec$
so, again using $S = u\,{t_{total}}$ we have,
$S = 12.2\sqrt {\dfrac{{490}}{{108}} \times \dfrac{{198}}{{490}}}$
$\Rightarrow S = 12.2 \times 1.35$
$\therefore S = 16.47\,m$
Hence, the ball will strike the ground at a distance of $S = 16.47m$ from the server.

Hence, the correct option is B.

Note: It should be remembered that, while falling under the force of gravity freely, the initial velocity of the body is always zero and acceleration is acceleration due to gravity and if a body is falling in downward direction acceleration due to gravity is always positive while if body is thrown upward its acceleration is taken as negative.