Question

A ball is projected from ground at an angle o with the horizontal. After 1 s, ball is moving at an angle 45° with the horizontal and after 3 s it is moving horizontally. The
velocity of projection of ball is (g = 10 m/s2)
0 10/13 m/s
0 2015 mis
01055 mis
010-70 mis​

Answer 1

Given :-

Angle of Projection = Ø

After 1 s it is moving with an angle of 45°.

tanØ = Vy/Vx

At time t = 3s its velocity in y-direction will become zero.

Vy = u SinØ - gt

0 = u SinØ - 3g

u SinØ = 3g -----(1)

Vx = u CosØ

tan 45° = u SinØ - gt/u CosØ

u CosØ = u SinØ - g

u CosØ = 2g ----(2)

Squaring and adding equation (1) and (2).

u² Sin²Ø + u² Cos²Ø = 9g² + 4g²

u²(Sin²Ø + Cos²Ø) = 13g²

u² = 13g²

u = √13g²

u = g√13

u = 10√13 ms-¹.

Hence,

The velocity of Projection of ball =u=10√13 ms-¹.