Q.A cricket ball is thrown at a speed of $28\ \text{m s}^{-1}$ in a direction $30^\circ$ above the horizontal. Calculate

Concept: Projectile Motion

(a) Maximum height ($H$)

At maximum height, the vertical component of velocity is zero. Using the kinematic equation $v_y^2 = u_y^2 + 2a_y \Delta y$:

$0 = (u \sin\theta)^2 - 2gH$ …

Projectile motion is a fundamental concept in kinematics, describing the path of an object thrown into the air and subject only to the acceleration of gravity. The key insight is that we can decompose the motion into two independent parts: horizontal and vertical.

The horizontal motion is uniform, meaning the horizontal velocity remains constant throughout the flight (assuming no air resistance). This is because gravity acts purely vertically, so there is no acceleration in the horizontal direction.

The vertical motion, however, is uniformly accelerated motion. The acceleration is due to gravity, acting downwards, which we typically denote as $g = 9.8\ \text{m s}^{-2}$. This constant downward acceleration causes the vertical velocity to change, first decreasing as the object rises, becoming zero at the peak of its trajectory, and then increasing in the downward direction as it falls.

By treating these two components separately, we can use the standard equations of motion (kinematic equations) for each direction.

Let's break down the problem:

1. Decompose Initial Velocity:

   The cricket ball is thrown with an initial speed $u = 28\ \text{m s}^{-1}$ at an angle $\theta = 30^\circ$ above the horizontal. We need to find its initial horizontal ($u_x$) and vertical ($u_y$) velocity components.

   • Horizontal component: $u_x = u \cos\theta$
   • Vertical component: $u_y = u \sin\theta$

   Substituting the given values:

   $u_x = 28 \cos 30^\circ = 28 \times \frac{\sqrt{3}}{2} = 14\sqrt{3}\ \text{m s}^{-1}$

   $u_y = 28 \sin 30^\circ = 28 \times \frac{1}{2} = 14\ \text{m s}^{-1}$

   For the vertical motion, the acceleration is $a_y = -g = -9.8\ \text{m s}^{-2}$ (taking upward as positive).

   For the horizontal motion, the acceleration is $a_x = 0$.

2. Calculate (a) the maximum height ($H$):

   The maximum height is reached when the vertical component of the ball's velocity momentarily becomes zero. At this point, the ball stops moving upwards before it starts falling downwards. We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement.

   For vertical motion: $v_y^2 = u_y^2 + 2a_y s_y$

   Here, $v_y = 0$ (at maximum height), $u_y = 14\ \text{m s}^{-1}$, $a_y = -9.8\ \text{m s}^{-2}$, and $s_y = H$.

   $0^2 = (14)^2 + 2(-9.8)H$

   $0 = 196 - 19.6H$

   $19.6H = 196$

   $H = \frac{196}{19.6} = 10\ \text{m}$

3. Calculate (b) the time taken by the ball to return to the same level (Time of Flight, $T$):

   The ball returns to the same level when its net vertical displacement from the starting point is zero. We can use the kinematic equation relating displacement, initial velocity, acceleration, and time.

   For vertical motion: $s_y = u_y t + \frac{1}{2} a_y t^2$

   Here, $s_y = 0$ (returns to the same level), $u_y = 14\ \text{m s}^{-1}$, $a_y = -9.8\ \text{m s}^{-2}$, and $t = T$.

   $0 = 14T + \frac{1}{2}(-9.8)T^2$

   $0 = 14T - 4.9T^2$

   We can factor out $T$:

   $T(14 - 4.9T) = 0$

   This gives two possible solutions for $T$: …

Concept: Derive the General Projectile Trio Once, Substitute Numbers Last

Method: Symbolic Derivation First (Energy for $H$, Symmetry for $T$, Then $R$), Numbers Only at the End

Rather than plugging $u=28$, $\theta=30^\circ$ into a fresh kinematic equation for each part, this method derives the three standard formulas $H(\theta)$, $T(\theta)$, $R(\theta)$ symbolically, using a different underlying principle for each, and substitutes the actual numbers only in the very last step.

Step 1 -- Maximum height $H$, via the work-energy theorem

At the highest point, only the vertical component of velocity has vanished; energy conservation applied to the vertical motion alone (the horizontal component $u\cos\theta$ is unchanged throughout and carries no energy loss/gain) gives:

Step 2 -- Time of flight $T$, via up-down symmetry

The vertical velocity is $v_y(t) = u\sin\theta - gt$. Time to reach the top ($v_y=0$): $t_{\text{up}} = \dfrac{u\sin\theta}{g}$. Because gravity is constant and the ball returns to the same launch height, the fall from the peak is the exact time-reverse of the rise -- so $t_{\text{down}}=t_{\text{up}}$, and:

Step 3 -- Range $R$, from horizontal motion during that time

Horizontal velocity never changes ($u\cos\theta$ throughout), so:

(using the double-angle identity $2\sin\theta\cos\theta=\sin2\theta$).

Step 4 -- Now substitute $u=28\ \text{m s}^{-1}$, $\theta=30^\circ$, $g=9.8\ \text{m s}^{-2}$ into all three at once

$T = \frac{2(28)\sin30^\circ}{9.8} = \frac{2(28)(0.5)}{9.8} = \frac{28}{9.8} = \frac{20}{7}\ \text{s}$ …