How Do You Calculate the Motion of a Baseball Thrown Upwards?

[gungo]

Homework Statement

A baseball is thrown straight up with a speed of 25 m/s. Determine:
(a) The maximum height reached by the ball
(b) How long it takes the ball to reach its maximum height
(c)The time it takes for the ball to reach a velocity of 5m/s (down)
(d) The speed of the ball when it returns to the same height from which it was thrown
(e) The time it takes for the ball to return to the same height from which it was thrown

Homework Equations

v2=v1+at
v2^2=v1^2+2ad

The Attempt at a Solution

I got the answers for everything but I'm nut sure if (e) is correct
(a)=31.89 m
(b)=2.55 s
(c)3.061 s
(d)25 m/s [down]
For e,
v2=-25m/s
v1=25 m/s
a=-9.8 m/s^2
-25=25+-9.8t
-50=-9.8t
t=5.1 s
Can someone just verify this? I have a huge test coming up, thanks!

 

[PeroK]

Looks fine. You might like to think about the relationship between the answers to b) and e).

 

[gungo]

Looks fine. You might like to think about the relationship between the answers to b) and e).

Oh so will the time it takes to reach the same height from which it was thrown always be double the time for the maximum height since it's halfway? Or are there exceptions?

 

[PeroK]

Oh so will the time it takes to reach the same height from which it was thrown always be double the time for the maximum height since it's halfway? Or are there exceptions?

Motion under constant gravity is symmetrical, so the downward motion is a mirror image of the upward motion.

In particular, it takes as long to fall back to the starting point as it does to rise to the highest point. If you want to set yourself a challenge, you could try to prove this using the kinematic equations.