A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20,10). If T is the time of flight, then its momentum vector, at time t=T/√2, is _____________.
[Take g=10 m / s² ]

Solution

Physics sir · 4 steps

Use the point on the path to find the launch speed first; after that it is just two velocity components.

1. Find the speed. The path equation is y = x tan θ - (g x²)/(2u²cos² θ). At 45°, tan θ = 1 and 2 cos² θ = 1, so it becomes y = x - (g x²)/u².

   10 = 20 - (10 × 20²)/u² ⇒ u² = 400, u = 20 m s⁻¹

2. Components and time. uₓ = u_y = 20 cos 45° = 10√2 m s⁻¹. The time of flight is T = 2u_y/g = 2√2 s, so t = T/√2 = 2 s.
3. Velocity then. The horizontal velocity stays 10√2 m s⁻¹. At t = 2 s, the vertical velocity is v_y = u_y - gt = (10√2 - 20) m s⁻¹.
4. So. Multiply by m = 10 kg: p⃗ = 100√2 î + (100√2 - 200) ĵ kg m s⁻¹, which is option (d).

The idea to keep. When a projectile passes through a given point, put that point into the trajectory equation to get u. Horizontal momentum never changes, and vertical momentum drops by mgt.

Answer from the NTA final answer key. Solution by PrepWiser.