Question

a body is projected horizontally with a velocity of 4m/s from top of a high tower. the velocity of the body after 0.7s is

Answer 1

Horizontal velocity after 0.7s
v_x = ut
= 4 * 0.7
= 2.8 m/s

Vertical velocity after 0.7s
v_y = gt
= 10 * 0.7
= 7 m/s

Resultant velocity after 0.7s
V = sqrt(v_x^2 + v_y^2)
= sqrt(2.8^2 + 7^2)
= 7.54 m/s

Answer 2

I hope after seeing this you will understand