Question

A ball is projected horizontally form the top of
a tower with velocity 4 m/s. The approx
velocity of the ball after 0.7 sec-(g = 10 m/s2)
(1) 11 m/s
(2) 10 m/s
(3) 8 m/s
(4) 3 m/s​

Answer 1

Answer:

8m/s

Explanation:

As there is no acceleration in the x direction so velocity in the x-direction remains constant=4m/s

As in the y-direction V=v+at. v(initial)=0

V=-10(0.7)= -7m/s.

(-) sign represents moving downwards

V(total)=√Vx^2 + Vy^2 = √ (4)^2 +(-7)^2 =√65= 8.06 m/s