Q.A cricket fielder can throw the cricket ball with a speed $v_o$. If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal, find

The key idea here is Projectile Motion combined with Relative Velocity. The ball's initial velocity relative to the ground is the vector sum of its velocity relative to the fielder and the fielder's velocity relative to the ground.

Let the velocity of the ball relative to the fielder be $\vec{v}_{bf}$ and the velocity of the fielder relative to the ground be $\vec{v}_{fg}$. The velocity of the ball relative to the ground is $\vec{v}_{bg} = \vec{v}_{bf} + \vec{v}_{fg}$.

Given:

• Speed of ball relative to fielder: $v_o$
• Angle of projection relative to horizontal (fielder's frame): $\theta$
• Speed of fielder: $u$ (horizontally)

The components of $\vec{v}_{bf}$ are $(v_o \cos\theta, v_o \sin\theta)$.

The components of $\vec{v}_{fg}$ are $(u, 0)$.

Thus, the components of $\vec{v}_{bg}$ are:

$v_{bg,x} = v_o \cos\theta + u$

$v_{bg,y} = v_o \sin\theta$

(a) The effective angle to the horizontal ($\alpha$) as seen by a spectator is given by the ratio of the vertical to horizontal components of the ball's velocity relative to the ground. …

When a fielder throws a ball while running, the ball's motion as seen by a spectator on the ground is a combination of two velocities: the velocity with which the fielder throws the ball relative to himself, and the velocity of the fielder relative to the ground. This is a classic problem involving relative velocity and projectile motion.

The key idea is to determine the initial velocity components of the ball relative to the ground. Once we have these, we can treat it as a standard projectile motion problem. The fielder is running horizontally, so his velocity only affects the horizontal component of the ball's velocity relative to the ground. The vertical component of the ball's velocity relative to the ground will be the same as its vertical component relative to the fielder.

Let's denote:

• $v_o$: Speed of the ball relative to the fielder.
• $u$: Speed of the fielder relative to the ground (horizontal).
• $\theta$: Angle at which the fielder throws the ball relative to the horizontal (as seen by the fielder).
• $g$: Acceleration due to gravity.

(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.

1. Resolve the ball's velocity relative to the fielder:

   The fielder throws the ball with speed $v_o$ at an angle $\theta$ to the horizontal.

   • Horizontal component of ball's velocity relative to fielder: $v_{ox} = v_o \cos \theta$
   • Vertical component of ball's velocity relative to fielder: $v_{oy} = v_o \sin \theta$

2. Determine the ball's velocity components relative to the ground:

   The fielder is running horizontally with speed $u$. This speed adds to the horizontal component of the ball's velocity. The vertical component remains unchanged.

   • Horizontal component of ball's velocity relative to ground ($V_x$): $V_x = v_{ox} + u = v_o \cos \theta + u$
   • Vertical component of ball's velocity relative to ground ($V_y$): $V_y = v_{oy} = v_o \sin \theta$

3. Calculate the effective angle:

   Let $\alpha$ be the effective angle to the horizontal at which the ball is projected in air as seen by a spectator. This angle is given by the ratio of the vertical to the horizontal components of the ball's velocity relative to the ground.

Therefore, the effective angle is $\boxed{\alpha = \arctan \left( \frac{v_o \sin \theta}{v_o \cos \theta + u} \right)}$.

(b) What will be the time of flight?

1. Identify the relevant velocity component:

   The time of flight of a projectile depends only on its initial vertical velocity component and the acceleration due to gravity. The horizontal motion does not affect the time the ball spends in the air.

   The initial vertical velocity of the ball relative to the ground is $V_y = v_o \sin \theta$.

2. Apply the time of flight formula:

   For a projectile launched from the ground and landing back on the ground, the time of flight $T$ is given by:

   $T = \frac{2 V_y}{g}$

   Substituting $V_y = v_o \sin \theta$:

The time of flight is $\boxed{T = \frac{2 v_o \sin \theta}{g}}$.

(c) What is the distance (horizontal range) from the point of projection at which the ball will land?

1. Recall the definition of horizontal range: The horizontal range $R$ is the product of the horizontal component of the ball's velocity relative to the ground and the total time of flight.

2. Substitute the expressions for $V_x$ and $T$: From part (a), $V_x = v_o \cos \theta + u$. From part (b), $T = \frac{2 v_o \sin \theta}{g}$.

Using the identity $2 \sin \theta \cos \theta = \sin(2\theta)$:

The horizontal range is $\boxed{R = \frac{v_o^2 \sin(2\theta) + 2 u v_o \sin \theta}{g}}$.

(d) Find $\theta$ at which he should throw the ball that would maximise the horizontal range as found in (c).

1. Set up for maximization: To find the angle $\theta$ that maximizes the range $R$, we need to differentiate $R$ with respect to $\theta$ and set the derivative equal to zero.

Set $\frac{dR}{d\theta} = 0$:

Divide by $2 v_o$ (assuming $v_o \neq 0$):

2. Solve the trigonometric equation: Use the double angle identity $\cos(2\theta) = 2 \cos^2 \theta - 1$:

This is a quadratic equation in terms of $\cos \theta$. Let $x = \cos \theta$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Since $\theta$ is an angle of projection, it must be between $0^\circ$ and $90^\circ$, so $\cos \theta$ must be positive. We take the positive root:

The angle $\theta$ for maximum range is $\boxed{\theta = \arccos \left( \frac{-u + \sqrt{u^2 + 8v_o^2}}{4v_o} \right)}$.

(e) How does $\theta$ for maximum range change if $u > v_o$, $u = v_o$, $u < v_o$?

Let's analyze the expression for $\cos \theta$:

1. Case 1: $u = 0$ (fielder is stationary)

This gives $\theta = 45^\circ$, which is the standard result for maximum range when there is no horizontal motion of the projection point.

2. Case 2: $u > 0$ (fielder is running)

We need to compare $\cos \theta$ with $\cos 45^\circ = \frac{1}{\sqrt{2}}$.

Let's rewrite the expression for $\cos \theta$:

Consider the term $\sqrt{u^2 + 8v_o^2}$. Since $u > 0$, the numerator is positive.
As $u$ increases, the term $\sqrt{u^2 + 8v_o^2}$ increases, but the $-u$ term also increases in magnitude.
Let's look at the derivative of $\cos \theta$ with respect to $u$:
$$ \frac{d(\cos \theta)}{du} = \frac{1}{4v_o} \left( -1 + \frac{2u}{2\sqrt{u^2 + 8v_o^2}} \right) = \frac{1}{4v_o} \left( -1 + \frac{u}{\sqrt{u^2 + 8v_o^2}} \right) $$ …

Concept: Keep the Range Formula in Product Form $R=2V_xV_y/g$ Instead of Expanding to $\sin2\theta$

Method: Differentiate the Product $V_x(\theta)V_y(\theta)$ Directly

The direct method expands the range into $\dfrac{v_0^2\sin2\theta+2uv_0\sin\theta}{g}$ and differentiates term by term. This method keeps the general range formula for any projectile in its compact product form, $R=\dfrac{2V_xV_y}{g}$, and differentiates the product $V_x(\theta)V_y(\theta)$ directly using the product rule.

Steps

1. Write the ground-frame velocity components (as in the direct method): $V_x(\theta)=u+v_0\cos\theta$, $V_y(\theta)=v_0\sin\theta$. (This answers part (a): $\alpha=\arctan(V_y/V_x)$.)

2. Recall the general range formula for any launch with components $(V_x,V_y)$ over flat ground: since time of flight is $T=2V_y/g$ (part b) and range is horizontal speed $\times$ time,

(This is part (c), left in product form rather than expanded.)

3. Differentiate the product $V_xV_y$ directly using the product rule, instead of expanding $R$ into a sum first:

4. Set the derivative to zero:

This is the same equation the direct (expand-first) method reaches, but arrived at via one product-rule step instead of expanding $\sin2\theta=2\sin\theta\cos\theta$ into the range formula first and then differentiating two separate terms.

5. Solve the resulting quadratic in $\cos\theta$ exactly as before: $2v_0\cos^2\theta+u\cos\theta-v_0=0 \Rightarrow \cos\theta = \dfrac{-u+\sqrt{u^2+8v_0^2}}{4v_0}$. …