a stone is from the top of a tower 300m high and at the same time another stone is projected vertically upward with initial velocity 75m/s. calculate when and where the two stone meet

Answer 1

d1 + d2 = 300m,
0.5gt^2 + (75t + 0.5gt^2) = 300m,

Since g in the 1st term is positive and
g in the 3rd is negative, they cancel
each other and the Eq becomes:

75t = 300,
t = 4 seconds.

d1 = 0.5*9.8*4^2 = 78.4m. below top
of tower = 300 - 78.4 = 222m above
ground.

d2 = 75*4 - 0.5*9.8*4^2,
d2 = 300 - 78.4 = 222m above ground.

Therefore, the 2 stones meet 4s after
release at a distance of 222m above
ground.

Answer 2

U suck in explanation