physic- free fall?

Question

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 18 s, then the motor stops. The rocket altitude 22 s after launch is 4700 m. You can ignore any effects of air resistance.
(a) What was the rocket's acceleration during the first 18 s?
(b) What is the rocket's speed as it passes through a cloud 4700 m above the ground?

Murtaza · Accepted Answer

a) 

First of all lets find its speed after the motor closes .
Time = 22- 18 = 4 s

s = ½at²+ut

s1 + s2 = 4700

½a(18)² + ½(-9.8)(4)² + u(4) = 4700 .
162a + 4u = 4778.4 

V = u + at
u = a(18)
u = 18a 

162a + 4(18a) = 4778.4
180a = 4778.4
a = 26.546666667 m/s²
u = 18(26.546666667)
u = 477.84 m/s

Therefore its speed after 18 is 477.84 m/s
Its acceleration is 26.546666667 .

Now speed at 4700 m

V = u + at 
V = (477.84) + (-9.8)(4)
V = 438.64 m/s

Anonymous · Answer

the answerer above had a mathematical error.
162a + 4(18a) = 234a (not 180a)

Vf = at + Vi

the velocity of the rocket after 18s is:
Vf = 18a + 0
Vf = 18a

the distance it goes during that 18s is:
Xf = .5a(18)Â²
Xf = 162a

after 18s, the rocket become a free-falling object, and starts to decelerate at the rate of -9.8 m/sÂ² during the next 4s.

recall that it travels at 18a m/s after 18s
the distance it goes during the next 4s is:
Xf = .5(-9.8)(4)Â² + (18a)(4) 
Xf = -78.4 + 72a

given that the altitude after 22s is 4700m. 
the distance it goes during 18s and 4s together add up to 4700m

so,
4700 = 162a - 78.4 + 72a
4778.4 = 234a
a = 20.42051282 m/sÂ² < answer to part A


so we know acceleration of the rocket during the first 18s. The velocity of the rocket after 18s is:
Vf = (20.42051282) (18)
Vf = 367.5692308 m/s

then after 18s, it decelerates at the rate of -9.8m/sÂ² for the next 4s

Vf = (-9.8)(4) + 367.5692308
Vf = 328.3692308 m/s <== answer to part B

physic- free fall?

2 Answers