Question

A body is projected vertically up from the foot of a tower 392 m height with a velocity 98 m/s at the same instant another body dropped from the top. Find when and where they will meet?

Answer 1

Sis the rightos Asky

Answer 2

Answer:

Let them meet at a height x above the ground t seconds after projection.

For the body A,

U = + 98 m/s, a = - 9.8m/s2

S= x, t = t

S = = ut + ½ at^2

x = 9.8 t – 4.9 t^2 ……………….(1)

For the second body B,

U = 0, a = 9.8 m/s^2

s = (392 – x), t = t

392 – x = 4.9 t^2 …………..(2)

Adding (1) and (2) 392 = 98t, t = 4 s

Substituting the value of t in equation (1)

h = 98 x 4 – 4.9 x 4 2 = 313.6 m