A body of mass m thrown up vertically with ve | vedclass

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

Question

(B) For the first body thrown vertically upwards with velocity $v_1$, the time taken to reach maximum height is $t_1 = \frac{v_1}{g}$ and the maximum

Vedclass · Accepted Answer

$4: 1$

Vedclass · Answer

(B) For the first body thrown vertically upwards with velocity $v_1$, the time taken to reach maximum height is $t_1 = \frac{v_1}{g}$ and the maximum height is $h_1 = \frac{v_1^2}{2g}$.For the second body projected with velocity $v_2$ at an angle $	heta$, the time taken to reach maximum height is $t_2 = \frac{v_2 \sin 	heta}{g}$ and the maximum height is $h_2 = \frac{v_2^2 \sin^2 	heta}{2g}$.Given $t_1 = 2t_2$, we have $\frac{v_1}{g} = 2 \left( \frac{v_2 \sin 	heta}{g} ight)$, which simplifies to $v_1 = 2 v_2 \sin 	heta$.Now, the ratio of heights is $\frac{h_1}{h_2} = \frac{v_1^2 / 2g}{v_2^2 \sin^2 	heta / 2g} = \frac{v_1^2}{v_2^2 \sin^2 	heta}$.Substituting $v_1 = 2 v_2 \sin 	heta$, we get $\frac{h_1}{h_2} = \frac{(2 v_2 \sin 	heta)^2}{v_2^2 \sin^2 	heta} = \frac{4 v_2^2 \sin^2 	heta}{v_2^2 \sin^2 	heta} = 4$.Thus, the ratio is $4:1$.

Similar Questions

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

When a player throws a ball,it reaches the other player in $4 \,s$. If the height of each player is $1.8 \,m$,the maximum height attained by the ball above the ground is (in $\,m$)

$A$ body projected from the ground reaches a point $X$ in its path after $3 \ s$ and from there it reaches the ground after a further $6 \ s$. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The acceleration due to gravity on this planet is ......... $m/s^2$.

$A$ stone is thrown vertically at a speed of $30 \,ms^{-1}$ making an angle of $45^{\circ}$ with the horizontal. What is the maximum height reached by the stone (in $\,m$)? Take $g=10 \,ms^{-2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module