Question

a stone is dropped from a tower 300 m high and at the same time another is projected vertically upward with velocity of 100 ms find when and where two stones meet ​

Answer 1

the ans is 3 sec let A be for tower and B for stone thrown vertically upward

Answer 2

Answer:

Explanation:

upward motion-

u = 100m/s g= -10

s=ut + 1/2 gt^2

s= 100t - 1/2 (10)t^2

s= 100t - 5t^2........1

downward motion

300- s = 100t + 5t^2.........2

adding (1) & (2)

300=200t

t=3/2s

from eqn. 1

s = 100*3/2-5*(3/2)^2

and answer will give u the height at which the stones will meet from the ground