Question

A car is moving horizontally along a straight line with a uniform velocity of 25$m{s^{ - 1}}$. A projectile is to be ruled from this car in such a way that it will return to it after it has moved $100m$. the speed of projectile must be:
A. 10$m{s^{ - 1}}$
B. 20$m{s^{ - 1}}$
C. 15$m{s^{ - 1}}$
D. 25$m{s^{ - 1}}$

Answer 1

Hint:Use the formula of the time of flight of projectile and the concept of vertical and horizontal components of a projectile to obtain the correct answer. The time taken by the projectile to return to the car will be the same as the time in which the car has moved $100m$.

Complete step by step answer:
A body that is thrown with some initial horizontal velocity or having a component along horizontal and then allowed to move under the effect of gravity alone without being propelled by an engine or fuel is called a projectile. Let the projectile be thrown out of the car at point A and it returns to the car after it has moved 100m. The velocity of the car is 25 $m{s^{ - 1}}$and it is moving along a straight line so the time taken by the car to reach point B from point A can be calculated as follow
$
velocity = \dfrac{{displacement}}{{time}} \\
\Rightarrow time = \dfrac{{displacement}}{{velocity}} \\
\Rightarrow time = \dfrac{{100}}{{25}}
\Rightarrow time = 4\sec \\ $
This is the same time that is taken by the projectile to return to the car.
So the time of flight = 4 seconds or,
$
\dfrac{{2u\operatorname{Sin} \theta }}{g} = 4 \\
\Rightarrow u\operatorname{Sin} \theta = 2g \\ $
Taking $g = 10m{s^{ - 2}}$
We get, $u\operatorname{Sin} \theta = 20m{s^{ - 1}}$
This is the vertical component of the speed with which the projectile is thrown. So this is the speed with which the projectile is launched vertically. And it will return to the car with the same speed of $20m{s^{ - 1}}$.

Hence, option B is the correct answer.

Additional information:
The horizontal range is the maximum distance traveled by the projectile during its time of flight, it is given as, $R = \dfrac{{{u^2}\operatorname{Sin} 2\theta }}{g}$.
For the maximum horizontal range, the projectile should be thrown at an angle of 45° with the plain.

Note:The value of g is taken as 10 instead of 9.8 to round off the answer. And air resistance is ignored in this answer. On breaking of a vector into its parts namely horizontal (x-axis) component and vertical (y-axis) component, these parts obtained are called its components. The horizontal component, uCosθ is supplied by the moving car and thus is equal to the velocity of the moving car, i.e. 25$m{s^{ - 1}}$.