Question

A projectile is projected with velocity of 40 m/s at an angle $\theta$ with the horizontal. If $R$ be the horizontal range covered by the projectile and after $t$ seconds, its inclination with horizontal becomes zero, then the value of $\cot \theta$ is [Take, $g=10$ m/s$^2$]

• A. $\frac{R}{20t^2}$
• B. $\frac{R}{10t^2}$
• C. $\frac{5R}{t^2}$
• D. $\frac{R}{t^2}$

Answer 1

Answer: (A)

$\frac{R}{20t^2}$

Answer 2

The projectile is projected with an initial velocity $u = 40$ m/s at an angle $\theta$ with the horizontal.

The velocity components are $u_x = u \cos \theta = 40 \cos \theta$ and $u_y = u \sin \theta = 40 \sin \theta$.

The acceleration due to gravity is $g = 10$ m/s$^2$ downwards.

The inclination of the projectile with the horizontal becomes zero at the highest point of its trajectory. The vertical component of velocity is zero at this point.

Let $t$ be the time taken to reach the highest point. The vertical velocity at time $t$ is given by $v_y(t) = u_y - gt$.

At $t$, $v_y(t) = 0$.

So, $0 = u \sin \theta - gt$

$u \sin \theta = gt$

Substituting $u = 40$ and $g = 10$:

$40 \sin \theta = 10t$

$\sin \theta = \frac{10t}{40} = \frac{t}{4}$ (Equation 1)

The horizontal range $R$ is the horizontal distance covered by the projectile when it returns to the initial height. The time taken for this is the total time of flight, which is twice the time taken to reach the highest point.

Total time of flight $T = 2t$.

The horizontal range is given by $R = u_x \times T$.

$R = (u \cos \theta) \times (2t)$

Substituting $u = 40$:

$R = (40 \cos \theta) \times (2t)$

$R = 80t \cos \theta$ (Equation 2)

We need to find the value of $\cot \theta$. We know that $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

From Equation 1, we have $\sin \theta = \frac{t}{4}$.

From Equation 2, we have $\cos \theta = \frac{R}{80t}$.

Now, we can find $\cot \theta$:

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{R}{80t}}{\frac{t}{4}}$

$\cot \theta = \frac{R}{80t} \times \frac{4}{t}$

$\cot \theta = \frac{4R}{80t^2}$

$\cot \theta = \frac{R}{20t^2}$