Question

A boy throws a ball in air at to${{60}^{0}}$ the horizontal along a road with a speed of 10m/s. Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car. If the car has a speed of (18 km/h). Give an explanation to support your diagram.

Answer 1

In this question the boy threw the ball at some angle. This means that the motion of the ball will be affected by gravity so the concept of projectile motion is implemented. When the motion of this ball is observed by a moving car, then the concept of relative velocity of car and ball has to be used for calculation.

Complete answer:
The ball moves in projectile motion when it is thrown at some angle.
Projectile Motion: When a particle is thrown obliquely at some angle near the earth’s surface, it moves along a curved path under constant acceleration that's directed towards the centre of the world (we assume that the particle remains on the brink of the surface of the earth). The path of such a particle is named a projectile and therefore the motion is named as projectile motion.
Two types of rectilinear motion in projectile motion are:
Along the x-axis of motion: uniform velocity, liable for the horizontal (forward) motion of the particle.
Along y-axis of motion: uniform acceleration, responsible for the vertical (downwards) motion of the particle.
Relative velocity: Relative velocity is defined by the fact that velocity of an object B with respect to the rest frame of another object A.
So, the velocity of ball is 10m/s
And the velocity of the car is 18 km/hr.
For finding the relative velocity of the ball with respect to the car, we shall use the concept of projectile motion.
Let us assume ${{u}_{x}}$be the horizontal velocity and ${{u}_{y}}$be the vertical velocity.

& {{u}_{x}}=10\cos \theta \\\ & {{u}_{x}}=(10m/s)\cos {{60}^{o}}=10\times \dfrac{1}{2}=5m/s \\\ & {{u}_{y}}=10\sin \theta \\\ & {{u}_{y}}=(10m/s)\sin {{60}^{o}}=10\times \dfrac{\sqrt{3}}{2}=5\sqrt{3}m/s \\\ \end{aligned}$$  This image shows that the ball is thrown at an angle of $${{60}^{o}}$$from the horizontal. So, the x-component of the ball is visible to the boy inside the car. Since speed of the car is $$18{}^{Km}/{}_{h}=5{}^{m}/{}_{s}$$ as the horizontal speed of the ball and car are the same, the relative velocity of the ball with respect to the car in the horizontal direction will be 0. Since the relative speed of the horizontal component is 0.So only vertical motion of the ball will be observed by the boy in the car, as shown in above diagram. **Note:** For projectile motion both components of velocity are taken together as one component is responsible for horizontal motion and the other is responsible for vertical motion. For relative velocity calculation if both are in the same direction then relative velocity is calculated by taking the difference between the velocities.