For a projectile, the ratio of maximum height | vedclass

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

Question

(a) $H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$ and $T = \frac{{2u\sin 	heta }}{g}$

So $\frac{H}{{{T^2}}} = \frac{{{u^2}{{\sin }^2}	heta /2g}}{{

Vedclass · Accepted Answer

$5:4$

Vedclass · Answer

(a) $H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$ and $T = \frac{{2u\sin 	heta }}{g}$

So $\frac{H}{{{T^2}}} = \frac{{{u^2}{{\sin }^2}	heta /2g}}{{4{u^2}{{\sin }^2}	heta /{g^2}}} = \frac{g}{8} = \frac{5}{4}$

For a projectile, the ratio of maximum height reached to the square of flight time is ($g = 10 ms^{-2}$)

Similar Questions

A particle is projected from ground at an angle $\theta$ with horizontal with speed $u$. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is ........

Two bodies are thrown up at angles of $45^o $ and $60^o $, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown is

For a given velocity, a projectile has the same range $R$ for two angles of projection if $t_1$ and $t_2$ are the times of flight in the two cases then

From the top of a tower of height $40\, m$, a ball is projected upwards with a speed of $20\, m/s$ at an angle $30^o$ to the horizontal. The ball will hit the ground in time ......... $\sec$ (Take $g = 10\, m/s^2$)

A particle projected from ground moves at angle $45^{\circ}$ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance]