Question 31: Motion in 2D & Vectors

Question

The angle of projection for a projectile to have same horizontal range and maximum height is :

Accepted Answer

To find the angle of projection for a projectile to have the same horizontal range and maximum height, we need to express both the range and the maximum height in terms of the projectile's initial velocity and the angle of projection, and then set them equal to each other.
The formula for the horizontal range $R$ of a projectile is given by:
$R = \frac{v^2}{g} \sin 2	heta,$
where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $	heta$ is the angle of projection.
The formula for the maximum height $H$ reached by the projectile is:
$H = \frac{v^2}{2g} \sin^2	heta.$
To have the same numerical value for $R$ and $H$, we set them equal to each other:
$\frac{v^2}{g} \sin 2	heta = \frac{v^2}{2g} \sin^2	heta.$
Simplifying this equation, we get:
$2 \sin 2	heta = \sin^2	heta.$
Using the double-angle formula, $\sin 2	heta = 2 \sin	heta \cos	heta$, we can rewrite the equation as:
$4 \sin	heta \cos	heta = \sin^2	heta.$
This can be simplified further to:
$4 \sin	heta \cos	heta = (\sin	heta)^2.$
Dividing both sides by $\sin	heta$ (assuming $\sin	heta 
eq 0$), we get:
$4 \cos	heta = \sin	heta.$
Now, dividing both sides by $\cos	heta$, we have:
$4 = 	an	heta.$
So, the angle of projection $	heta$ is:
$	heta = 	an^{-1}(4).$
Therefore, the correct answer is:
Option D $$	an ^{-1}(4)$$.

Answer

$$	an ^{-1}\left(\frac{1}{2}ight)$$

Answer

$$	an ^{-1}(2)$$

Answer

$$	an ^{-1}\left(\frac{1}{4}ight)$$

Answer

$$	an ^{-1}(4)$$

Motion in 2D & Vectors

The angle of projection for a projectile to have same horizontal range and maximum height is :

Choose the correct answer: