A projectile thrown with velocity v making an | vedclass

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

Question

$\begin{array}{l}
Max.\,height = H = \frac{{{v^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}}\,\,\,....\left( i ight)\
Time\,of\,flight,\,T =

Vedclass · Accepted Answer

$\sqrt {8H/g} $

Vedclass · Answer

$\begin{array}{l}
Max.\,height = H = \frac{{{v^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}}\,\,\,....\left( i ight)\
Time\,of\,flight,\,T = \frac{{2\,v\,\sin \left( {90 - 	heta } ight)}}{g}\,\,\,\,\,...\left( {ii} ight)\
From\left( i ight),\,\frac{{v\,\cos \,	heta }}{g} = \sqrt {\frac{{2H}}{g}} \,From\,\left( {ii} ight),\
T = 2\sqrt {\frac{{2H}}{g}}  = \sqrt {\frac{{8H}}{g}} .
\end{array}$

A projectile thrown with velocity $v$ making angle $\theta$ with vertical gains maximum height $H$ in the time for which the projectile remains in air, the time period is

Similar Questions

A projectile crosses two walls of equal height $H$ symmetrically as shown The angle of projection of the projectile is

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

A projectile is thrown with velocity $u$ making angle $\theta$ with vertical. It just crosses the tops of two poles each of height $h$ after $1\,s$ and $3\,s$, respectively. The maximum height of projectile is ............ $m$

Two bodies are projected with the same velocity. If one is projected at an angle of ${30^o}$ and the other at an angle of ${60^o}$ to the horizontal, the ratio of the maximum heights reached is

The maximum height reached by a projectile is $64 \mathrm{~m}$. If the initial velocity is halved, the new maximum height of the projectile is_________.$\mathrm{m}$.