# Thread: Problem

1. ## Problem

A rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 16 seconds. At this time it has reached a height of 560 metres above the launch pad and attained an upward velocity of 90ms-1. From this time on, the rocket has constant upward acceleration -10ms-2 (due to the effect of gravity alone).
Choose the s-axis (for the position of the particle that represents the rocket) to point upwards, with origin at the launch pad. Take t = 0 to be the time when the rocket motor cuts out.

(a) What is the maximum height (above the launch pad) reached by the rocket?

v²-2as = v0²-2as0

where a = -10ms-2, s0= 0 and v0 = 90

Thus when v= 0, we obtain

s= v02 / 2a = 90²/2x10 = 405

So the maximum height reached by the rocket is 405 metres above the launch pad

b) How long (from launch) does the rocket take to reach this maximum height?

t= v-v0/a

v= 10t; a= -10; v0= 90

= 10-90/-10 = 8

It take 8 seconds for the rocket to reach the maximum height.

(c) After how long (from launch) does the rocket crash onto the launch pad?

where a = -10ms-2, s0= 405 and v0 = 90
s= 1/2 at² + v0t + s0
= -102/2 + 90t + 405
= -5t2 + 90t + 405 = 0

Using the quadratic formula which give t= -9 and t= 12.7

I have difficulties with the last question, I don't think it's correct
Can someone help me

2. Originally Posted by valerie-petit
A rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 16 seconds. At this time it has reached a height of 560 metres above the launch pad and attained an upward velocity of 90ms-1. From this time on, the rocket has constant upward acceleration -10ms-2 (due to the effect of gravity alone).
Choose the s-axis (for the position of the particle that represents the rocket) to point upwards, with origin at the launch pad. Take t = 0 to be the time when the rocket motor cuts out.

(a) What is the maximum height (above the launch pad) reached by the rocket?

v²-2as = v0²-2as0

where a = -10ms-2, s0= 0 and v0 = 90

Thus when v= 0, we obtain

s= v02 / 2a = 90²/2x10 = 405

So the maximum height reached by the rocket is 405 metres above the launch pad

b) How long (from launch) does the rocket take to reach this maximum height?

t= v-v0/a

v= 10t; a= -10; v0= 90

= 10-90/-10 = 8

It take 8 seconds for the rocket to reach the maximum height.

(c) After how long (from launch) does the rocket crash onto the launch pad?

where a = -10ms-2, s0= 405 and v0 = 90
s= 1/2 at² + v0t + s0
= -102/2 + 90t + 405
= -5t2 + 90t + 405 = 0

Using the quadratic formula which give t= -9 and t= 12.7

I have difficulties with the last question, I don't think it's correct
Can someone help me
(c) Find time taken from max height to hitting ground (take DOWNWARDS as positive direction):

t = T = ?
initial velocity = 0 m/s
displacement = 405 m
acceleration = 10 m/s^2 (since downwards is positive).

Total time = 8 + T ....

(For checking purposes I get T = 9 seconds).

3. Do I have to calculate my quadratic equation with initial velocity = 0?

Sorry, I'm sure it is easy problem, but really I can't see how it work

4. Never mind I got it