Motion in a Plane — Physics STD 11 Science — Question

1. At maximum height v =0

$\therefore\ 0=\text{u}\sin\theta-\text{gt};$

or $\text{t}=\frac{\text{u}\sin\theta}{\text{g}}$

But time of flight,

$\text{T}=\text{2t}=\frac{2\text{u}\sin\theta}{\text{g}}$

2. When the body returns to the same horizontal level y = 0

$\therefore 0=\text{x}\tan\theta-\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta}$ [From (i)]

or $\text{x}\tan\theta=\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta}$

or $\text{x}=\frac{2\text{u}^2\sin\theta\cos\theta}{\text{g}}=\frac{\text{u}^2\sin2\theta}{\text{g}}$

But coordinates of M are (R, 0).

Putting x = R, we have $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$