Question

A stone is thrown vertically into the air from the top of tower 100m high. At the same instant a second stone is thrown upward from the ground. The initial velocity of first stone is 50m/s and that of second stone is 75m/s. When and where will the

stones be at the same height from the ground.​

Answer 1

Answer:

The stones meet after 4 seconds at a height of

121.6 m from the bottom to the tower

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Answer 2

Concept:

The second equation of motion,

s = ut + (1/2) at²

where s is the distance traveled, u is the initial velocity, a is acceleration and t is the time.

Given:

The height of the tower, h = 100m

The velocity of the first stone, v₁ = 50 m/s

The velocity of the second stone, v₂ = 75 m/s

Find:

The height from the ground and the time at which both stones will meet.

Solution:

Let, the distance traveled by the first stone is x and distance traveled by the first stone be y.

According to the equation of motion,

s = ut + (1/2)at²

Distance traveled by the first stone, acceleration is equal to the acceleration due to gravity as the stone is thrown downwards,

x = 50t + (1/2)gt²

Similarly, distance traveled by the second stone, acceleration is equal to the acceleration due to gravity with a negative sign as the stone is thrown upwards,

y = 75t + (1/2)-gt² =  75t - (1/2)gt²

The stones will meet somewhere between the tower,

So, x + y = h  (Height of tower) = 100 m

Substituting the values of x and y,

x + y = 50t + (1/2)gt² + 75t - (1/2)gt² = 100

125 t = 100 m

t = 100/125 s = 0.8 s

Distance from the ground at the point they will meet will be y at t = 0.8 s,

y =  75t - (1/2)gt² =  75 (0.8)  - (1/2)×9.8(0.8)²

y = 60 - 3.136 = 56.864 m

Hence, the stones will meet at a distance of 56.864 m from the ground after 0.8 s.

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