A rocket is fired vertically up from the grou | vedclass

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

Question

(A) The distance travelled by the rocket in $1\ min$ $(= 60\ s)$ during which the resultant acceleration is vertically upwards and $10\ m/s^2$ is give

Vedclass · Accepted Answer

$36\ km, 1\ min$

Vedclass · Answer

(A) The distance travelled by the rocket in $1\ min$ $(= 60\ s)$ during which the resultant acceleration is vertically upwards and $10\ m/s^2$ is given by $h_1 = \frac{1}{2} 	imes 10 	imes 60^2 = 18000\ m = 18\ km$. The velocity acquired by the rocket at the end of $1\ min$ is $v = u + at = 0 + 10 	imes 60 = 600\ m/s$. After the fuel is finished, the rocket moves vertically up under gravity. The acceleration is $-g = -10\ m/s^2$. It reaches a maximum height $h_2$ when its final velocity becomes $0$. Using $v^2 = u^2 + 2as$, we get $0 = (600)^2 - 2 	imes 10 	imes h_2$, which gives $h_2 = \frac{360000}{20} = 18000\ m = 18\ km$. The total maximum height reached from the ground is $h = h_1 + h_2 = 18\ km + 18\ km = 36\ km$. To find the time taken after the fuel is finished to reach the maximum height, we use $v = u + at$, where $v = 0$, $u = 600\ m/s$, and $a = -10\ m/s^2$. Thus, $0 = 600 - 10t$, which gives $t = 60\ s = 1\ min$.

Similar Questions

Gravels are dropped on a conveyor belt at the rate of $0.5 \, kg/s$. The extra force required in newtons to keep the belt moving at $2 \, m/s$ is

$A$ rocket with a lift-off mass $20,000\; kg$ is blasted upwards with an initial acceleration of $5.0\; m s^{-2}$. Calculate the initial thrust (force) of the blast.

$A$ rocket is fired vertically from the ground with a resultant vertical acceleration of $10 \,m/s^2$. The fuel is exhausted in $1 \,min$ and it continues to move up. What is the maximum height reached (in $\,km$)?

$A$ balloon has a mass of $10\,g$ in air. The air escapes from the balloon at a uniform rate with a velocity of $4.5\,cm/s$. If the balloon shrinks completely in $5\,s$,then the average force acting on the balloon will be (in $dyne$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module