A projectile is fired at an angle of 45^{circ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the angle of projection be $	heta = 45^{\circ}$.At the highest point $P$, the height is $H = \frac{u^2 \sin^2 	heta}{2g}$ and the horizontal

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the angle of projection be $	heta = 45^{\circ}$.At the highest point $P$, the height is $H = \frac{u^2 \sin^2 	heta}{2g}$ and the horizontal distance from the point of projection $O$ is $R/2 = \frac{u^2 \sin 	heta \cos 	heta}{g}$.The elevation angle $\alpha$ is the angle subtended by the highest point $P$ at the point of projection $O$ with the horizontal.In the right-angled triangle $	riangle POM$, we have:$	an \alpha = \frac{H}{R/2} = \frac{\frac{u^2 \sin^2 	heta}{2g}}{\frac{u^2 \sin 	heta \cos 	heta}{g}} = \frac{\sin 	heta}{2 \cos 	heta} = \frac{1}{2} 	an 	heta$.Substituting $	heta = 45^{\circ}$:$	an \alpha = \frac{1}{2} 	an 45^{\circ} = \frac{1}{2} (1) = \frac{1}{2}$.Therefore, $\alpha = 	an^{-1}\left(\frac{1}{2}ight)$.

$A$ projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

Similar Questions

$A$ particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$ it is away from the target by $10 \ m$, then the angle of projection to hit the target is

The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

$A$ ball is thrown at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. This is possible only when the value of $\tan \theta$ is ..........

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight, then the value of maximum range is (acceleration due to gravity $= g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module