Two projectiles,one fired from the surface of | vedclass

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

Question

(A) For two projectiles to trace identical trajectories,their maximum heights $(H)$ and horizontal ranges $(R)$ must be equal.The formula for maximum

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) For two projectiles to trace identical trajectories,their maximum heights $(H)$ and horizontal ranges $(R)$ must be equal.The formula for maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$.Since the trajectories are identical,the angle of projection $	heta$ is the same for both.Equating the heights: $\frac{u_1^2 \sin^2 	heta}{2g_1} = \frac{u_2^2 \sin^2 	heta}{2g_2}$.Substituting the given values ($u_1 = 10 \, m/s$,$u_2 = 5 \, m/s$,and $g_1 = 10 \, m/s^2$):$\frac{10^2}{2 	imes 10} = \frac{5^2}{2 	imes g_2}$.$\frac{100}{20} = \frac{25}{2g_2}$.$5 = \frac{12.5}{g_2}$.$g_2 = \frac{12.5}{5} = 2.5 \, m/s^2$.

Two projectiles,one fired from the surface of the Earth with velocity $10 \, m/s$ and the other fired from the surface of another planet with initial speed $5 \, m/s$,trace identical trajectories. The value of the acceleration due to gravity on the planet is ......... $m/s^2$.

Similar Questions

An object is projected from the ground with a speed of $20 \, m/s$ at an angle of $30^{\circ}$ with the horizontal. Its centripetal acceleration one second after the projection is .......... $m/s^2$ [Take $g = 10 \, m/s^2$].

$A$ ball $P$ is dropped vertically and another ball $Q$ is thrown horizontally with the same initial vertical velocity (which is $0$) from the same height $h$ at the same time. If air resistance is neglected,then:

$A$ body is projected with a velocity $u$ at an angle $\theta$ with the horizontal. If the time of ascent of the body is $1 \ s$,then the maximum height it can reach is (Take $g = 10 \ m/s^2$) (in $m$)

$A$ body is projected at an angle of $30^{\circ}$ with the horizontal with momentum $p$. At its highest point,the magnitude of the momentum is

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module