How Does SUVAT Calculate Displacement and Maximum Height of a Thrown Ball?

[Peter G.]

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.

 

[gneill]

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?

 

[Peter G.]

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]

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

 

[Peter G.]

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]

Some kinematics equations that you might find helpful:

d(t) = d₀ + v₀*t + (1/2)at²

v(t) = v₀ + a*t

 

[Peter G.]

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

 

[gneill]

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.