Question

A body is projected horizontally with a
velocity of 20m/s from the top of a tower of
height 490m. The distance at which it falls
from the top of the tower is​

Answer 1

Answer:

Explanation:

sy= -490m

uy= 0

ay=-10m/s^2

t=?

s=ut+1/2at^2

t=$\sqrt{98}$s=9.9s or 10s

sx=x

ux=20m/s

ax=0

t=10s (approx.)

s=ut+1/2at^2

x=200m (approx.)