Question

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

Answer 1

hope u understand yourself

Answer 2

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 + $\frac{1}{2}at^2$

so,

x = (0) t + $\frac{1}{2}gt^2$ ...(i)

The stone that is thrown vertically upwards,

u = -50 m/s^-1

s = -(200 - x)

a = g

by using the formula s = ut + $\frac{1}{2}at^2$

so,

−(200−x) = −50 * t + $\frac{1}{2}gt^2$

⇒ 200 = 50t −  $\frac{1}{2}gt^2$ + x    ....(ii)

By solving the equations (i) and (ii)

200 = 50t + $\frac{1}{2}gt^2$ - $\frac{1}{2}gt^2$

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

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