Question

from the top of a tower height 78.4 two stones are projected horizontally with 20 metre per second and 30 metre per second in opposite direction are reaching the ground Their Separation is​

Answer 1

Given data : From the top of a tower height 78.4 metre two stones are projected horizontally with 20 metre per second and 30 metre per second in opposite direction are reaching the ground.

To find : Sepration of the stones ?

Solution : Let, initial velocity ( u ) of stones be zero.

Here, we use formula of kinemetical equation : s = u * t + ½ * a * t2

Where,

• s is the didplacement
• u is initial the velocity
• t is time taken by the particle
• a is acceleration of the particle

Let, displacement of both stones is 78.4 metre and a be the acceleration due to gravity, hence, a = g and accelerayltion due to gravity is 9.8 m/s².

Now,

⟹ s = u * t + ½ * a * t²

⟹ 78.4 = 0 * t + ½ * 9.8 * t²

⟹ 78.4 = 4.9 * t²

⟹ t² = 78.4/4.9

⟹ t² = 16

⟹ t = √16

⟹ t = 4 sec

Let, speed of first stone be 20 m/s and speed of second stone be 30 m/s

Now, by formula of speed :

⟹ speed of first stone = distance/time

⟹ 20 = distance/4

⟹ distance = 20 * 4

⟹ distance = 80 m

Hence, the distance covered by first stone is 80 m

⟹ speed of second stone = distance/time

⟹ 30 = distance/4

⟹ distance = 30 * 4

⟹ distance = 120 m

Hence, the distance covered by second stone is 120 m.

⟹ sepration of the stone = addition of distance covered by first stone and distance covered by second stone.

⟹ sepration of the stone = 80 + 120

⟹ sepration of the stone = 200 m

Answer : Hence, sepration between two stones is 200 m.

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