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

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

Explanation:

Given that,

Initial velocity

Altitude = 1494 m

We need to calculate the distance

Using equation of motion

Where, v = final velocity

u = initial velocity

a = acceleration due to gravity

s = distance

Put the value into the formula

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

The distance traveled by the rocket

We need to calculate the time

Using formula of time

Where, v = speed

t = time

d = distance

Put the value into the formula

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

lexy is right because she has the labeled s

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6