The horizontal range and maximum height attai | vedclass

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

Question

(D) The vertical motion of the projectile is independent of the horizontal motion. Since the vertical acceleration remains $-g$,the maximum height $H$

Vedclass · Accepted Answer

$(R+H), H$

Vedclass · Answer

(D) The vertical motion of the projectile is independent of the horizontal motion. Since the vertical acceleration remains $-g$,the maximum height $H$ remains unchanged.For the horizontal range,the initial range is $R = u_x T$,where $T = \frac{2 u_y}{g}$ is the time of flight.The new horizontal range $R^{\prime}$ with horizontal acceleration $a_x = \frac{g}{4}$ is given by:$R^{\prime} = u_x T + \frac{1}{2} a_x T^2$Substituting $R = u_x T$ and $T^2 = \frac{4 u_y^2}{g^2}$:$R^{\prime} = R + \frac{1}{2} \left(\frac{g}{4}\right) \left(\frac{4 u_y^2}{g^2}\right)$Since $H = \frac{u_y^2}{2g}$,we have $u_y^2 = 2gH$. Substituting this:$R^{\prime} = R + \frac{g}{8} \left(\frac{4(2gH)}{g^2}\right) = R + \frac{g}{8} \left(\frac{8gH}{g^2}\right) = R + H$Thus,the new range is $(R+H)$ and the height is $H$.

The horizontal range and maximum height attained by a projectile are $R$ and $H$,respectively. If a constant horizontal acceleration $a=g/4$ is imparted to the projectile due to wind,then its horizontal range and maximum height will be

Similar Questions

Three projectiles $A, B$ and $C$ are projected at an angle of $30^{\circ}, 45^{\circ}, 60^{\circ}$ respectively. If $R_{A}, R_{B}$ and $R_{C}$ are the ranges of $A, B$ and $C$ respectively,then (velocity of projection is the same for $A, B$ and $C$):

$A$ cannon on a level plane is aimed at an angle $\theta$ above the horizontal and a shell is fired with a muzzle velocity $v_0$ towards a vertical cliff a distance $D$ away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

$A$ body is projected from the ground at an angle of $\tan^{-1}(\sqrt{7})$ with the horizontal. At half of the maximum height, the speed of the body is '$n$' times the speed of projection. The value of '$n$' is

$A$ boy throws a ball with a velocity $u$ at an angle $\theta$ with the horizontal. At the same instant,he starts running with a uniform velocity to catch the ball before it hits the ground. To achieve this,he should run with a velocity of

Two stones having different masses $m_1$ and $m_2$ are projected at an angle $\alpha$ and $(90^{\circ}-\alpha)$ with the same speed from the same point. The ratio of their maximum heights is

Vedclass Test Series

Exam Paper Generator

Online Exam Module