# SUVAT - Ball Thrown upwards

A Ball is thrown vertically upwards at 20 m/s, find its displacement after:
a) 1 second
b) 5 seconds
c) What is the maximum height the ball can reach

(Take the acceleration due to gravity to be of 10 m/s)

a) I did: s = ut + 0.5 x at^2
I got 15 m, which I'm pretty sure is correct.

b) For this one, I first tried: s = (20 x 5) + (0.5 x -10 x 5^2) and I got a displacement of 25 m, which, sounds a bit strange since the ball would be traveling at 20 m/s by the fourth second and would hit the ground at 30 m/s.

So I tried the same equation but using the initial velocity 0, and therefore, instead of five, three seconds. With the maths I got 45 m displacement.

Which one is right please so I know how to approach other problems like these better?

And for C, as I looked at it straight away without thinking or doing any maths I thought that the ball would reach a height of 30 m, because for the first second it would move 20 m and for the next second, it would move 10 m (20 m/s decelerated by gravity to 10 m/s). I mean I know it is wrong, but I simply can't get my head around it.

What is getting me confused regarding that is that the ball is only moving at 20 m/s in the very first moment after that, it is already decelerating, but one second later, it would be in fact moving at 10 m/s, right?

Thanks,
Peter.

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gneill
Mentor
Watch the sign on displacement; is it 25m or -25m?

For part c, when the ball is at its maximum height, what do you expect the velocity to be?

Oh, for the displacement he didn't give us any specific instruction so I used bearings, which we were using previously, so for a) I gave the displacement, 15 m, bearing 0 and for b) 25, or 45 Bearing 180.

Well, for c, the ball will start at 20 m/s, gradually decelerate 19, 18, 17 until 10 m/s when the first second goes by and then gradually decrease speed until when the second second strikes and it shows 0 m/s.

gneill
Mentor
So you've determined the time when it's at its zenith. How far does it travel in that time?

I think I got it for C: It takes two seconds to reach its maximum height, that is when it has no speed. Then, to figure out the height, or distance (S) we have enough information, such as u = 20, v = 0, a = -10 and t = 2.

But I am in doubt with B. Both equations seem reasonable, but they give different results. I'm pretty sure the right answer is 45 m at Bearing 180 (I also drew a distant time graph and it also indicated the same)

gneill
Mentor
Some kinematics equations that you might find helpful:

d(t) = d0 + v0*t + (1/2)at2

v(t) = v0 + a*t

Nice, never seen/used them before. Why next to the d and v there is a 0 subscript?

gneill
Mentor
Nice, never seen/used them before. Why next to the d and v there is a 0 subscript?
It just shorthand for the initial value. Typically it's the value at t = 0.