Solving Linear Motion Problems: Initial Speed, Time, and Height

[MooPhysics]

Homework Statement

"A ball thrown straight up with an initial speed of 30/m/s.
(a) Show the time it takes to reach its trajectory will be 3m/s
(b) Show that it will reach a height of 45m"

Homework Equations

h=ut+1/2g^2
1/2gt^2

The Attempt at a Solution

Part a of the question I am fine with.
Its part (b) that seems to be what got me confused. I felt the need to use the equation "h=ut+1/2g^2" due to thinking that: u=30? Initial speed in this case is 30m/s. Why can't I use it in the equation? When and what type of questions would I use h=ut+1/2g^2??
I know that to work this question out, I will need to use "1/2gt^2" I just don't understand why initial speed (u) won't be 30m/s.

Also questions related to physics. How do you know when to work out -in relation to velocity or acceleration? I know the difference between the two, yet seem to get confused in exams/tests between them both? Is there a trick into isolating the questions into what they are asking rather than me assuming what they are asking?

 

[PeroK]

The initial speed is $30m/s$. But, these equations involve velocities, not speeds. So, you need to get the right sign +/- for your velocity and acceleration.

Hint: The ball is thrown up but gravity is a downward force.

 

[MooPhysics]

The initial speed is $30m/s$. But, these equations involve velocities, not speeds. So, you need to get the right sign +/- for your velocity and acceleration.

Hint: The ball is thrown up but gravity is a downward force.

I used 1/2gt^2, Since knowing that time is 3 seconds I worked it out as: 1/2x10x3^2=45m.
I used a positive gravity (10-rounded up 9.81 for a solid number), and used 3seconds due to it being half the total time, and therefore the time it would have taken to reach maximum height. Is this line of thinking correct? In what case can I expect to use the equation- h=ut+1/2g^2 ??

 

[PeroK]

I used 1/2gt^2, Since knowing that time is 3 seconds I worked it out as: 1/2x10x3^2=45m.
I used a positive gravity (10-rounded up 9.81 for a solid number), and used 3seconds due to it being half the total time, and therefore the time it would have taken to reach maximum height. Is this line of thinking correct?In what case can I expect to use the equation- h=ut+1/2g^2 ??

In this case you got the right answer by accident! You should use that last formula in this case.