For a projectile,the ratio of the maximum hei | vedclass

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

Question

(A) The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.We need to

Vedclass · Accepted Answer

$5 : 4$

Vedclass · Answer

(A) The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.We need to find the ratio $\frac{H}{T^2}$.$\frac{H}{T^2} = \frac{\frac{u^2 \sin^2 	heta}{2g}}{(\frac{2u \sin 	heta}{g})^2} = \frac{u^2 \sin^2 	heta}{2g} 	imes \frac{g^2}{4u^2 \sin^2 	heta}$.Simplifying the expression,we get $\frac{H}{T^2} = \frac{g}{8}$.Taking $g = 10 \ m/s^2$,the ratio is $\frac{10}{8} = \frac{5}{4}$.

For a projectile,the ratio of the maximum height reached to the square of the time of flight is:

Similar Questions

$A$ ball can be thrown to a maximum horizontal distance of $100\,m$. To what maximum height can it be thrown?

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

If the initial kinetic energy of a projectile is $100 \, J$ and the kinetic energy at the maximum height is $30 \, J$,what is the angle of projection?

The range of a projectile for a given initial velocity is maximum when the angle of projection is $45^\circ$. The range will be minimum,if the angle of projection is ......... $^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module