a stone is dropped from the top of building 200m hai and at the same time another stone is projected vertically upward from ground with velocity of 50 M per second find where and when the two stones will meet
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a stone is dropped from the top of building 200m hai and at the same time another stone is projected vertically upward from ground with velocity of 50 M per second find where and when the two stones will meet
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The stones will meet after 4 sec at the height of 120m.
Explanation:
Given that,
Height of the building = 200 m
The Velocity of the stone thrown vertically upward = 50 m/s
Let the place where the two stones meet be x and the time taken to meet is t. We will consider the downward direction as +ve.
The stone that is dropped;
Its initial velocity u = 0 ms^-1
Acceleration = acceleration due to gravity(g)
Displacement (s) = x
by using the formula s = ut +
so,
x = (0) t + ...(i)
The stone that is thrown vertically upwards,
u = -50 m/s^-1
s = -(200 - x)
a = g
by using the formula s = ut +
so,
−(200−x) = −50 * t +
⇒ 200 = 50t − + x ....(ii)
By solving the equations (i) and (ii)
200 = 50t + -
t = 200/50
∵ t = 4 sec
now,
x = 200 - 1/2 * 10 * 4^2
x = 200 - 80
∵ x = 120 m
Thus, the stones will meet after 4 sec at the height of 120m.
Learn more: find the velocity
brainly.in/question/27294078