Motion in a Plane - Projectile Motion

A cricket ball is thrown with a velocity of $28 \text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level. (Take $g = 9.8 \text{ m/s}^2$)

Given: $u = 28 \text{ m/s}$, $\theta = 30^\circ$, $g = 9.8 \text{ m/s}^2$. (a) Maximum height: $H = \frac{u^2 \sin^2 \theta}{2g} = \frac{28^2 \times \sin^2 30^\circ}{2 \times 9.8} = \frac{784 \times 0.25}{19.6} = 10 \text{ m}$. (b) Time of flight: $T = \frac{2u \sin \theta}{g} = \frac{2 \times 28 \times \sin 30^\circ}{9.8} = \frac{56 \times 0.5}{9.8} \approx 2.86 \text{ s}$. (c) Horizontal range: $R = \frac{u^2 \sin 2\theta}{g} = \frac{28^2 \times \sin 60^\circ}{9.8} = \frac{784 \times 0.866}{9.8} \approx 69.3 \text{ m}$.

We use the standard formulae for projectile motion. The maximum height depends on the vertical component of velocity, while the range depends on both components and the total time the object stays in the air.

Show that for a projectile, the angle of projection $\theta$ for which the horizontal range and the maximum height are equal is given by $\theta = \tan^{-1}(4)$.

Condition: $R = H$. $\frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$ $\frac{2u^2 \sin \theta \cos \theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$ $2 \cos \theta = \frac{\sin \theta}{2}$ $4 = \frac{\sin \theta}{\cos \theta}$ $\tan \theta = 4 \implies \theta = \tan^{-1}(4)$.

By equating the expressions for Horizontal Range ($R$) and Maximum Height ($H$), we simplify the trigonometric terms to find the specific ratio of $\sin \theta$ to $\cos \theta$.