Calculating Vertical Motion: Am I on the Right Track?

[anthonyk2013]

Wondering if I am 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

 

[adjacent]

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

 

[anthonyk2013]

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

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?

 

[haruspex]

(b)

s=ut+1/2at²

s=400*40.77+1/2*(-9.81)*40.77²

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 realized that meant you'd made a mistake, but instead you just changed the sign.)

 

[anthonyk2013]

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 realized that meant you'd made a mistake, but instead you just changed the sign.)

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

 

[adjacent]

I should use V²=U²+2as

V²=400²+2*(-9.81)*1.8

V²=square root of 159964.684

V=399.95m/s

1.8 was given in kilo meters.You should use SI units in the Kinematics equations.
You have used m/s for speed m/s^2 for acceleration and a wrong distance.Convert it to meters first.