Question

6.70. A stone is projected upwards from the foot of a towar 50
m high with a velocity of 25 m/sec and at the same time an
other stone is dropped from the top of the tower. The two
stones cross each other after​

Answer 1

Answer:

Explanation:

ball A :        

s= 50m                                                                                              

u =0m/s

g= 9.8 m/s^2          

  ball B:

s= 50m                                                    

u = 25m/s

g= - 9.8 m/s^2          

let the distance from top to the meeting point be x . Then from down to meeting point is (50-x)

x = ut + 1/2gt^2= 0 + 1/2gt^2 = 1/2gt^2.................(i)

(50-x)= ut - 1/2gt^2 = 25t- 1/2gt^2...........................(ii)

On adding (i) & (ii), we get

x + 50- x =       1/2gt^2 +     25t- 1/2gt^2

50= 25t

t = 50/25 =2s