Two balls are projected simultaneously, one from the top and the other from the bottom of a building as shown below. The ball does not bounce back after hitting the ground. Find the time after which ball A and ball B collide.
↑ 30 m/s (Ball B from bottom)
20 m (height of building)
↓ 40 m/s (Ball A from top)
Answer (2)
To find the time after which ball A (projected from the top) and ball B (projected from the bottom) collide, we can set up equations for the motion of each ball. We'll assume the upward direction is positive.
Given:
Ball A is projected downward from the top of the building (20 m above the ground) with an initial velocity of 40 m/s.
Ball B is projected upward from the bottom of the building with an initial velocity of 30 m/s.
We'll determine their positions as functions of time t and find when these positions are equal.
Equation of motion for Ball A:
The position y A ( t ) of Ball A at time t is given by: y A ( t ) = 20 − 40 t − 2 1 g t 2
where g = 9.8 m/s 2 is the acceleration due to gravity.
Simplifying further: y A ( t ) = 20 − 40 t − 4.9 t 2
Equation of motion for Ball B:
The position y B ( t ) of Ball B at time t is given by: y B ( t ) = 30 t − 2 1 g t 2
Simplifying further: y B ( t ) = 30 t − 4.9 t 2
Equating positions to find collision time
The balls collide when y A ( t ) = y B ( t ) :
20 − 40 t − 4.9 t 2 = 30 t − 4.9 t 2
By simplifying, we find: 20 = 70 t
Solving for t : t = 70 20 = 7 2 seconds
Therefore, the balls collide 7 2 seconds or approximately 0.286 seconds after they are projected.
This solution involves understanding kinematic equations and solving them step-by-step to find the desired time of collision.
The balls collide approximately 0.286 seconds after being projected. Ball A is projected downward from the top with an initial velocity of 40 m/s, while Ball B is projected upward from the bottom with an initial velocity of 30 m/s, allowing us to determine the collision time using kinematic equations. By equating their positions and solving, we find the time of collision.
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