Question
A body is dropped from top of a cliff 240 m high. At the same time another body is thrown vertically upwards from the ground with a speed 48m/s. Find when and where the two balls meet.
Solution
Given a ball is dropped from the top of a cliff.
Initial position of ball, X01 = 240m
Initial velocity of ball, u1 = 0
Acceleration, a1 = g= -10 m/s2
Therefore, the position of the ball at any instant is,
When the ball is projected from the ground.
Initial position of ball, X02 = 0
Initial velocity of ball, u2 = 48m/s
Acceleration, a2 = g= -10 m/s2
Therefore, the position of the ball at any instant is given by
When the two balls will meet, the position of both the balls will be the same.
i.e., x1 = x2
t = 5s
Putting t = 5 in equation (1), we have
x1 =240-5t2
=240-5(5)
=115m
Thus, both the balls meat at t = 5s at a height of 115m from the ground.
Initial position of ball, X01 = 240m
Initial velocity of ball, u1 = 0
Acceleration, a1 = g= -10 m/s2
Therefore, the position of the ball at any instant is,
When the ball is projected from the ground.
Initial position of ball, X02 = 0
Initial velocity of ball, u2 = 48m/s
Acceleration, a2 = g= -10 m/s2
Therefore, the position of the ball at any instant is given by
When the two balls will meet, the position of both the balls will be the same.
i.e., x1 = x2
t = 5s
Putting t = 5 in equation (1), we have
x1 =240-5t2
=240-5(5)
=115m
Thus, both the balls meat at t = 5s at a height of 115m from the ground.
