Question

A ball is projected horizontally with a velocity of $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ from the top of a building $19.6m$ high. The distance of the point from the bottom of the building where the ball will hit the ground is $\left(g=9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)$

• A. $8m$
• B. $6.6m$
• C. $20.6m$
• D. $14.6m$

Answer 1

The correct option is A

$8m$

Step 1. Given data:

1. Height of the building = $19.6m$
2. Since the ball is projected horizontally, the component of velocity in the horizontal direction $V_x=4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$,
3. And its initial vertical velocity, the component of velocity in the vertical direction $V_y=0$
4. So the time required to strike the ground will be free-fall time.
5. Acceleration due to gravity = $9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Step 2. Formula used:

1. Time of flight, $t=\sqrt{\frac{2h}{g}}$, Where $h=$height of the building, $g=$acceleration due to gravity.
2. Also $$\begin{gathered} S_x=u_x.t+\frac{1}{2}a{t}^2 \\ {S}_x=u_x.t \end{gathered}$$, Where $S_x=$horizontal distance in the x-direction.,$a=0$ for the horizontal direction.

Step 3. Calculations:

Putting all the known values in the above equations, we get

$t=\sqrt{\frac{2\times 19.6}{9.8}}=2s$

Hence, $$\begin{gathered} S_x=u_x.t \\ {S}_x=4\times 2=8m \end{gathered}$$,

Thus, the distance of the point from the bottom of the building where the ball will hit the ground is $8m$.

Hence, option A is the correct option.