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

Let the stones meet at point A after time t.

For upper stone  :            

u′ =0      

x=0+  1/2 gt²

x=  1/2 ×10×t   ²

⟹ x= 5t²   ............(1)

For lower stone :            

u=25  m/s  

100−x=ut−  1/2gt²

100−x=(25)t− 1/2×10×t²

⟹ 100−x=25t−5t²   ............(2)

Adding (1) and (2),  we get          

25t=100              

⟹ t=4 s

From (1),            

x=5×4²

⟹x=80 m  

Hence the stone meet at a height of   20 m above the ground after 4 seconds.

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.