If R be the horizontal range of a projectile and h its greatest height, prove that its initial velocity is 2g (h + R^216h) . [6] OR A bullet passes through two planks in succession. Its initial…

Solution to the Projectile Motion Problem

We are to prove that the initial velocity $u$ of a projectile is given by: $$u=\sqrt{2g\left(h+\frac{R^2}{16h}\right)}$$ where:

• $R$ = horizontal range of the projectile,
• $h$ = maximum height attained by the projectile,
• $g$ = acceleration due to gravity.

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Step 1: Recall the equations of projectile motion

For a projectile launched with initial velocity $u$ at an angle $\theta$ with the horizontal, the following equations hold:

1. Maximum height ($h$): $$h=\frac{u^2{\sin }^2\theta }{2g}$$ This is derived from the vertical component of motion, where the final vertical velocity at the highest point is zero.

2. Horizontal range ($R$): $$R=\frac{u^2\sin 2\theta }{g}$$ This comes from the total horizontal distance traveled before the projectile hits the ground.

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Step 2: Express $\sin \theta$ and $\cos \theta$ in terms of $h$ and $R$

From the maximum height equation: $${\sin }^2\theta =\frac{2gh}{u^2}\Rightarrow \sin \theta =\sqrt{\frac{2gh}{u^2}}$$

From the range equation, using the double-angle identity $\sin 2\theta =2\sin \theta \cos \theta$: $$R=\frac{u^2\cdot 2\sin \theta \cos \theta }{g}\Rightarrow \sin \theta \cos \theta =\frac{Rg}{2u^2}$$

Square both sides: $${\sin }^2\theta {\cos }^2\theta ={\left(\frac{Rg}{2u^2}\right)}^2$$

Substitute ${\sin }^2\theta =\frac{2gh}{u^2}$: $$\left(\frac{2gh}{u^2}\right){\cos }^2\theta =\frac{R^2{g}^2}{4u^4}$$

Solve for ${\cos }^2\theta$: $${\cos }^2\theta =\frac{R^2{g}^2}{4u^4}\cdot \frac{u^2}{2gh}=\frac{R^{2}g}{8h{u}^2}$$

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Step 3: Use the Pythagorean identity

We know: $${\sin }^2\theta +{\cos }^2\theta =1$$ Substitute the expressions for ${\sin }^2\theta$ and ${\cos }^2\theta$: $$\frac{2gh}{u^2}+\frac{R^{2}g}{8h{u}^2}=1$$

Factor out $\frac{g}{u^2}$: $$\frac{g}{u^2}\left(2h+\frac{R^2}{8h}\right)=1$$

Solve for $u^2$: $$u^2=g\left(2h+\frac{R^2}{8h}\right)$$

Simplify the expression inside the parentheses: $$u^2=2gh+\frac{R^{2}g}{8h}$$

Factor out $2g$: $$u^2=2g\left(h+\frac{R^2}{16h}\right)$$

Take the square root of both sides to find $u$: $$u=\sqrt{2g\left(h+\frac{R^2}{16h}\right)}$$

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Solution to the Bullet and Planks Problem

Given:

• Initial velocity of the bullet, $u=1200\text{m/s}$.
• Velocity lost in penetrating each plank, $\Delta v=200\text{m/s}$.
• Let the thickness of the first plank be $t_1$ and the second plank be $t_2$.
• Both planks offer the same average resistance.

Assumption: The resistance offered by each plank is proportional to the thickness of the plank (i.e., the work done to penetrate the plank is proportional to its thickness).

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Step 1: Velocity after penetrating the first plank

After passing through the first plank, the velocity becomes: $$v_1=u-\Delta v=1200-200=1000\text{m/s}$$

Step 2: Velocity after penetrating the second plank

After passing through the second plank, the velocity becomes: $$v_2=v_1-\Delta v=1000-200=800\text{m/s}$$

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Step 3: Relate thickness to velocity loss

The work done to penetrate a plank is equal to the change in kinetic energy. Since the resistance is the same for both planks, the work done per unit thickness is constant.

The kinetic energy lost in penetrating the first plank: $$\Delta K{E}_1=\frac{1}{2}m{u}^2-\frac{1}{2}m{v}_1^2=\frac{1}{2}m(1200^2-1000^2)$$ $$\Delta K{E}_1=\frac{1}{2}m(1,440,000-1,000,000)=\frac{1}{2}m(440,000)=220,000m$$

The kinetic energy lost in penetrating the second plank: $$\Delta K{E}_2=\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_2^2=\frac{1}{2}m(1000^2-800^2)$$ $$\Delta K{E}_2=\frac{1}{2}m(1,000,000-640,000)=\frac{1}{2}m(360,000)=180,000m$$

Since the resistance per unit thickness is the same, the ratio of the thicknesses is equal to the ratio of the kinetic energy losses: $$\frac{t_1}{t_2}=\frac{\Delta K{E}_1}{\Delta K{E}_2}=\frac{220,000}{180,000}=\frac{11}{9}$$

Thus, the ratio of the thickness of the first plank to the second plank is: $$\boxed{\frac{11}{9}}$$