JEE Main 2021 · 26 February · evening · Single correct · +4 / −1
The trajectory of a projectile in a vertical plane is y = αx - βx², where α and β are constants and x y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by :
- (a) tan ⁻¹α ,(α ²)/4β
- (b) tan ⁻¹α ,(4α ²)/β
- (c) tan ⁻¹( β/α ),(α ²)/β
- (d) tan ⁻¹β ,(α ²)/2β
Answer and solution
Answer (a) tan ⁻¹α ,(α ²)/4β
Solution
Physics sir · 3 steps
The slope at the start gives the launch angle, and zero slope at the top gives the height.
- Angle of projection. The velocity at launch is along the tangent to the path at the origin. So tan θ equals the slope dy/dx = α - 2β x at x = 0, which is α. That gives θ = tan⁻¹ α.
- Find the top. At the highest point the path is flat, so set dy/dx = 0. That gives α - 2β x = 0, so x = α/2β.
H = α(α/2β) - β(α/2β)² = α²/2β - α²/4β = α²/4β
- So. θ = tan⁻¹ α and H = α²/4β, which is option (a).
The idea to keep. For any trajectory y = f(x), the slope at x = 0 is tan θ, and the maximum height is where the slope becomes zero. No need to find u or g separately.
Answer from the NTA final answer key. Solution by PrepWiser.