Question

A stone is dropped from the top of a tower of height 78.4 m. At the same time another stone is projected vertically up from the foot of the tower with a velocity 9.8ms^-1 . When the two stones will meet?​

Answer 1

Given:-

For The First stone:-

• Initial Velocity of Stone = 0m/s

• Acceleration Due to Gravity = + 9.8m/s²

• Height of the Tower = 78.4m

For The second Stone:-

• Initial Velocity = 9.8m/s¹

• Acceleration due to Gravity = -9.8m/s²

Formulae Used:-

• S = ut + ½ × a × t²

Where,.

S = Distance

u = Initial Velocity

a = Acceleration

t = Time

Now,

First We will find time taken by First stone to cover the full Distance.

S = ut + ½ × a × t²

78.4 = (0) × t + ½ × 9.8 × t²

78.4 = 0 + 4.9t²

78.4 = 4.9t²

t² = 78.4/4.9

t² = 16

√t² = √16

t = 4s.

Hence, For first and second stone ,t = 4s.

Now, we will find the distance covered by second stone, That will be our answer.

S = ut + ½ × a × t²

S = 9.8 × 4 + ½ × -9.8 × 4²

S = 39.2 + (-4.9 × 16)

S = 39.2 - 78.4

S = -39.2m

But, Distance can not be in negative so, When the First stone will cover 39.2m from the top he will meet Second stone.