Physics: Asmall rocket is launched vertically, attaining a maximum speed at burnout of 1.0 ✕ 10^2 m/s and thereafter coasting straight up to a maximum altitude of 1494 m. assuming the rocket accelerated uniformly while the engine was on, how long did it fire and how high was it at engine cut off

Asmall rocket is launched vertically, attaining

syrai7404

11.03.2022

The distance traveled and time taken by the rocket are 1488.89 m and 14.8 sec.

Explanation:

Given that,

Initial velocity $u= 1.0\times10^2\ m/s$

Altitude = 1494 m

We need to calculate the distance

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

a = acceleration due to gravity

s = distance

Put the value into the formula

$0=(1.0\times10^2)^2+2\times(-9.8)\times s$

$s=\dfrac{1.0\times10^{2}}{2\times9.8}$

$s=5.10204\ m$

So, The distance from the start of the coast to the end is 5.10204 m

The distance traveled by the rocket

$1494-5.10204=1488.89\ m$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Where, v = speed

t = time

d = distance

Put the value into the formula

$t=\dfrac{1488.89}{1.0\times10^{2}}$

$t=14.8\ sec$

Hence, The distance traveled and time taken by the rocket are 1488.89 m and 14.8 sec.

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