Question

A girl throws a water filled balloon on the eve of holi with a velocity 12 $m{s^{ - 1}}$ at an angle of ${60^0}$ with the horizontal. It hits a car at the same height (of throw) approaching her at 15$m{s^{ - 1}}$. Find the distance of the car from her.
A. 13 m
B. 31 m
C. 44 m
D.18 m

Answer 1

Hint: In order to solve this, as per the above question when a girl throws a balloon filled with water it makes some angle with horizontal. Hence we need to calculate the horizontal range of projection, time of flight. Then we can find the distance of the car.

Complete step by step answer:
When the balloon projected, the time taken by the balloon to reach surface is given by
$
t = \dfrac{{2V\sin \theta }}{g} \\
\Rightarrow t = 2.12s \\
$
 Horizontal range of projection by the balloon is given by
$
R = \dfrac{{{V^2}\sin 2\theta }}{g} \\
\Rightarrow R = 12.72m \\
$
The distance traveled by the car is given by
$ D = R \times t \\
\Rightarrow D =2.12×15 \\
\therefore D =31.8 m \\
$
Now, Distance of car from girl=12.72+31.8 ≅ 44 m.Hence the correct option is C.

Additional information:
When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the center of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.

Note:The horizontal range can be stated as the horizontal distance travelled by a body which is projected from its initial distance is at zero to the position where until its fall. The time of flight of a projectile motion is the total time taken by the object projected to the time taken to reach the earth’s surface.