rrect Option is (B)

Here, two projectiles are projected at angles 42${}^{\circ }$ and 48${}^{\circ }$ with same initial velocity.

As we know the expression of range of projectile,

Range $=\frac{u^2\sin 2\theta }{g}$

AT $\theta$₁ = 42${}^{\circ }$,

Range, $R_1=\frac{u^2\sin 2(42{)}^{\circ }}{g}=\frac{0.99u^2}{g}$

At $\theta$₂ = 48${}^{\circ }$

Range, $R_2=\frac{u^2\sin 2(48{)}^{\circ }}{g}=\frac{0.99u^2}{g}$

The range of the projectile is same for the two projectiles.

Therefore, R₁ = R₂

Now, as we know the expression of height of the projectile,

$H_{max}=\frac{u^2\sin \theta }{2g}$

At 42${}^{\circ }$, $H_{max}=H_1=\frac{u^2\sin {42}^{\circ }}{2g}=\frac{0.669u^2}{2g}$

At 48${}^{\circ }$, $H_{max}=H_2=\frac{u^2\sin {48}^{\circ }}{2g}=\frac{0.743u^2}{2g}$

Higher the value of $\theta$ higher the value of maximum height. Therefore, H₁ < H₂.

rrect Option is (C)

$R=\sqrt{\frac{2h}{g}}.v$

$D=\sqrt{R^2+h^2}$

$=\sqrt{{\left(\sqrt{\frac{2h}{g}}.v\right)}^2+h^2}$

$D=\sqrt{\frac{2h{v}^2}{g}+h^2}$

Option (c) is correct.

rrect Option is (C)

$H=\frac{U^2{\sin }^2\theta }{2g}$

$=\frac{{(25)}^2.{(\sin 45)}^2}{2\times 10}$

= 15.625 m

$T=\frac{U\sin \theta }{g}$

$=\frac{25\times \sin {45}^{\circ }}{10}$

= 2.5 $\times$ 0.7

= 1.77 s

rrect Option is (C)

Relative velocity of bomb w.r.t. observer in plane = 0.

Bomb will fall down vertically. So, it will move in straight line w.r.t. observer.

rrect Option is (A)

y = $\alpha$x $-$ $\beta$x²

comparing with trajectory equation

$y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2{\cos }^2\theta }$

$\tan \theta =\alpha \Rightarrow \theta ={\tan }^{-1}\alpha$

$\beta =\frac{1}{2}\frac{g}{u^2{\cos }^2\theta }$

$\Rightarrow$ $u^2=\frac{g}{2\beta {\cos }^2\theta }$

Maximum height H :

$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{g}{2\beta {\cos }^2\theta }\frac{{\sin }^2\theta }{2g}$

$\Rightarrow$ $H=\frac{{\tan }^2\theta }{4\beta }=\frac{{\alpha }^2}{4\beta }$