Question

17. 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:

31.175m , 2.623sec.

Explanation:

let the two stones meet at distance H from the ground and time t sec. after projection , and let g = 10 m/s²

1st. stone : 100 - H = gt²/2

2nd. stone :  H = 25t - gt²/2

100 = 25t + 5t²

5t² + 25t - 100 = 0

t²+ 5t - 20 = 0

t = {-5 ±√105}/2 = {√105 -5 } /2 = 2.623sec.

H = 25×2.623 - 5×2.623² = 31.175m

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.