Question

A man throws a shot at an angle of 45 with the horizontal plane from a height of 1.5 metres. If the shot strikes the ground at a horizontal distance of 30 metres, find the velocity of throw....(g=10 m per sec sq)​

Answer 1

A man throws a shot at an angle of 45° with the horizontal plane from a height of 1.5 m. if the shot strikes the ground at a horizontal distance of 30m.

To find : The initial velocity of the shot.

solution : let initial velocity of the shot is u.

vertical component of u, $u_y$ = usin45°

initial position, $y_0$ = 1.5 m

final position, y = 0

using formula, $y=y_0+u_yt+\frac{1}{2}a_yt^2$

⇒0 = 1.5 + usin45° t - 1/2 × 10 × t²

⇒0 = 1.5 + usin45°t - 5t²

⇒5t² - usin45°t - 1.5 = 0.......(1)

now horizontal component of u, $u_x$ = ucos45°

initial position, $x_0=0$

final position, x = 30m

using formula, $x=x_0+u_xt+\frac{1}{2}a_xt^2$

⇒30 = ucos45°t + 0

⇒30/ucos45° = t .......(2)

putting equation (2) in equation (1) we get,

5 × 900/u²cos²45° - usin45° × 30/ucos45° - 1.5 = 0

⇒9000/u² - 30 - 1.5 = 0

⇒9000/u² = 31.5

⇒9000/31.5 = u²

⇒u² = 285.7

⇒u = 16.9 m/s

Therefore initial velocity of the shot is 16.9 m/s