A stone is dropped from a height of 100 m, while another one is projected vertically upwards from the ground with a velocity of 25 m/s at the same time. The time in seconds after which they will have the same height is (acceleration due to gravity, `g = 10 ms^(-2)` )

Options

A. 4
B. 5
C. 6
D. 7

Answer

A

Explanation

Let stones meet after time t seconds.
Then distance travelled by first stone in t seconds =100-h(m) and distance travelled by IInd stone in t second is h (m). So, we have
For first stone, `-(100-h) = (-1)/(2) "gt"^2` ...(i)
For second stone `h = 25 t - 1/2 "gt"^2` .....(ii)
Substituting for `1/2 "gt"^2` from Eq. (i) in Eq. (i), we get
` h = 25t - (100 - h) rArr 25t = 100 `
` t= 4s`