Question

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.​

Answer 1

Answer:

t= 4s, x = 21.6m

For dropped stone -

X = 0+ 1/2gt2

For thrown up stine-

100-x = 25t-1/2gt2

from above two equations

get t= 4 s

put t =4 in first equation

get x= 78.4 m

100-x = 100-78.4= 21.6m

Answer 2

Answer:

let "t" = time after which both stones meet

"S" = distance travelled by the stone dropped from the top of tower

(100-S) = distance travelled by the projected stone.

◆ i) For stone dropped from the top of tower

-S = 0 + 1/2 (-10) t²

or, S = 5t²

◆ ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t²

= 25t - 5t²

Adding i) and ii) , We get

100 = 25t

or t = 4 s

Therefore, Two stones will meet after 4 s.

◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get

S = 5 × 16

= 80 m.

Thus , both the stone will meet at a distance of 80 m from the top of tower.