The trajectory of projectile, projected from the ground is given by y=x-x²/20. Where x and y are measured in meter. The maximum height attained by the projectile will be.

Solution

Physics sir · 4 steps

The top of the path is where the curve stops rising, so set dy/dx to zero and read off y.

1. Find the top. At maximum height the path is momentarily horizontal, so its slope is zero. Differentiate the trajectory and set the slope to zero.

   dy/dx = 1 - x/10 = 0 ⇒ x = 10 m

2. Put x back. y = 10 - 10²/20 = 10 - 5 = 5 m.
3. Quick check. Compare with y = x tan θ - (g x²)/(2u²cos² θ). Here tan θ = 1, so θ = 45°, and g/u² = 1/20 gives u² = 200 m² s⁻² with g = 10 m s⁻². Then H = (u²sin² θ)/2g = (200 × 1/2)/20 = 5 m.
4. So. The maximum height is 5 m, option (b).

The idea to keep. Given the trajectory equation, maximum height is the peak of y(x): set dy/dx = 0 to get x, then put it back to get y. The peak sits at half the range.

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