A stone is dropped from the top of a tower 400 m high and at the same time other stone is projected upward vertically from the ground with a velocity of 100 ms -1. Find where and when the two stones will meet. (a = 10 ms -2). (S)
Answers
Answer:
time = 4s
Distance =320m
Explanation:
Let the two stones meet each other at a distance x metre away and down from the top of the tower.
Consider the stone dropped from the tower.
Initial velocity, u = 0.
Acceleration, a = g = 10 m s^(-2)
At the point of meeting, which is x metres away from the top of the tower, the displacement of the stone dropped is x and is given by the second kinematic equation,
s = u t + (a t²) / 2,
i.e.,
x = 0 · t + (10 t²) / 2
x = 5 t² → (1)
where t is the time of meeting.
Now, consider the stone projected vertically upwards.
Initial velocity, u = 100 m s^(-1)
Acceleration, a = - g = - 10 m s^(-2)
At the point of meeting, which is actually (400 - x) metres high from the bottom of the tower, since the height of the tower is 400 m, the displacement of the stone projected is (400 - x) metres and is also given by the second kinematic equation, i.e.,
400 - x = 100 t + (- 10 t²) / 2
400 - x = 100 t - 5 t²
400 - x = 100 t - x [From (1)]
400 = 100 t
t = 4 s
So both the stones meet each other after 4 seconds of projection.
And the distance will be, from (1),
x = 5 (4)² = 80 m,
from the top, i.e., 400 - 80 = 320 metres from the bottom.