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Problem 1: A person throws a ball straight up with a speed of 12 m/s. If the bus is moving at 25 m/s, what is the velocity of the ball to an observer on the ground?

Solution

Theory: The ball when thrown, moves under the action of two type of velocities; one of the bus and the other its own velocity. The person in the bus goes with the ball with the same velocity as that of the bus and therefore, he feels only the velocity with which the ball is thrown. However, our observer on the road sees the ball moving in a different direction under the action of both velocities.  The resultant velocity would be the sum of both velocities. Therefore, we draw the velocity vectors to add them by head-to-tail rule and find the resultant.

Consider the diagram above. Let Vx is the velocity of the bus and Vy is the velocity with which the ball is thrown. We choose a scale such that 1 cm = 10 m/s. Therefore,

Vx = 2.5 cm = AB and Vy = 1.25 cm (approximately) = BC

By head-to-tail rule of vector addition, the resultant velocity is AC = 2.773 cm

Convert back the vector to velocity, 2.773 × 10 = 27.73 ≅ 28 m/s

To find the direction, apply Pythagoras theorem to the right-angled triangle,

Therefore, for an observer on the road, the ball seems to move with a velocity of 28 m/s making an angle of 26° with the horizontal direction.

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