Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DHard2 minQB

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Passage / Comprehension

A student designed a rocket that was launched straight up into the air from the ground. At $t = 0$ , the rocket is at $y = 0$ with $V_{y} (t = 0) = 0$ . The velocity of the rocket is given by: $V_{y} = (24 t - 3 t^{2}) m / s$ for $0 \leq t \leq t_{b}$ , where $t_{b}$ is the time at which the fuel burns out. The vertically upward direction is taken as positive. (Assume $g = 10 m / s^{2}$ )

The time taken for rocket to reach its maximum height is

Options:

Answer:

Solution:

for maximum height, $v_{y} = 0$ at $t = \frac{17}{3} s, v_{y} = 24 (\frac{17}{3}) - 3 (\frac{17}{3}) = \frac{119}{3} m / s$ after that,

$a = - g = - 10 m / s^{2}$

and at

$t = t_{2}, v_{y} = 0$

$0 = \frac{119}{3} - 10 (t_{2} - \frac{17}{3})$

or $t_{2} = \frac{289}{30} = 9.6 sec .]$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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