Question

A stone is projected at an angle of 45° above the
horizontal with speed 10√2 m/s from the top of a
building of height 200 m. If the point of projection
is taken as origin then the position
of the particle after 3.s of projection is (take
g= 10 m/s2)
​

Answer 1

Explanation:

v in x direction of ball = 10root2× cos45° = 10m/s= v in y direction.

Therefore, x distance= 10×3 = 30m

y distance= vt-1/2×g×t²

= 10×3 - 5×9 = -15m

Therefore, (30,-15)