# Linear Motion problem

1. Feb 26, 2014

### anthonyk2013

Wondering if im on the right track with below question

A body is projected vertically upwards from ground level with an initial velocity of 400m/s. determine;
(a) Time to reach the ground
(b) Velocity at a point 1.8km above the ground
(c) Time to reach a vertical height of 2.5km.

(a)
u=440m/s
v=0m/s
a=-9.81m/s

v=u+at
0=400+(-9.81)t
-400/-9.81=t
t=40.77sec

total time to reach ground 40.77*2=81.54sec

(b)

s=ut+1/2at2

s=400*40.77+1/2*(-9.81)*40.772

s=1908 + (-8153.05)

s=-6245.05 m = 6.24km

2. Feb 26, 2014

Where's c? And why did you find the distance for b when it asked for the velocity?

3. Feb 26, 2014

### anthonyk2013

only half way through (b) haven't started (c)

though I need the distance to calculate the velocity at 1.8km? maybe I m wrong?

4. Feb 26, 2014

### haruspex

Why are you using half the time from part (a)? That will give you the maximum height, no? The question does not ask for that.
You are given initial speed, acceleration, distance, and you are asked for final speed. What SUVAT equation connects those four?
(Btw, you dropped a factor of 10 in the first term of the calculation, giving you a negative maximum height. You should have realised that meant you'd made a mistake, but instead you just changed the sign.)

5. Feb 26, 2014

### anthonyk2013

I should use V2=U2+2as

V2=4002+2*(-9.81)*1.8

V2=square root of 159964.684

V=399.95m/s

(c)
Having trouble transposing formula to find time for part (C) not sure if the below is correct?
S=2.5kn or 2500m
u=400m/s
v=0m/s

S=(u+v)/2*t

t=s/(u+v)/2

t=2500/(400+0)/2

t=12.5sec

Last edited: Feb 26, 2014
6. Feb 27, 2014