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## Homework Statement

A particle is projected vertically upwards at 30 m/s. Calculate (a) how long it takes to reach its maximum height, (b) the two times at which it is 40 m above the point of projection, (c) the two times at which it is moving at 15 m/s.

**2. Homework Equations [/b**

final velocity (v)

initial velocity (u)

acceleration due to gravity (a) = 10m/s^2

time (t)

Distance or displacement (s)

v^2=u^2 + 2as

v=u+at

s=ut+1/2 at^2

s=1/2(u+v)t

I first find the maximum height, this is where i used the equation v^2=u^2 +2as

0^2=30^2 + 2(10)s

s=45 m maximum height

then i used that value and plugged into the equation to find time

45= 1/2 (30+0)t

t= 3seconds

therefore total time is 3 seconds for the particle to reach its maximum height.

part (b) is where i am stuck

however i move on to part (c)

and i plugged in the values in the equation

15=30-10t

t=1.5 seconds

however the question asked for two (2) times. so i am also stuck with finding the second time...

final velocity (v)

initial velocity (u)

acceleration due to gravity (a) = 10m/s^2

time (t)

Distance or displacement (s)

v^2=u^2 + 2as

v=u+at

s=ut+1/2 at^2

s=1/2(u+v)t

## The Attempt at a Solution

I first find the maximum height, this is where i used the equation v^2=u^2 +2as

0^2=30^2 + 2(10)s

s=45 m maximum height

then i used that value and plugged into the equation to find time

45= 1/2 (30+0)t

t= 3seconds

therefore total time is 3 seconds for the particle to reach its maximum height.

part (b) is where i am stuck

however i move on to part (c)

and i plugged in the values in the equation

15=30-10t

t=1.5 seconds

however the question asked for two (2) times. so i am also stuck with finding the second time...