Constant acceleration formula question

Nathan hits a tennis ball straight up into the air from a height of 1.25 m above the ground.

The ball hits the ground after 2.5 seconds. Air resistance is neglected.

the first question is to I find the speed with which Nathan hits the ball which is 11.75 m per second.

The second question is to find the greatest height above the ground reached by the ball Which is 8.29m.

the third question is to then find the speed of the ball when it hits the ground. This is the question I'm having trouble with; I know the answer and the correct pathway to get to the answer, I just don't know why the method outlined below is incorrect - please help me with this 🙂

S = 0 is defined at ground level.

Correct path is with v = u + at

Wrong path is with s = vt - 0.5at², although the right formula to use is s = vt - 0.5at² + 1.25 because when t = 0, s = 1.25

I know I haven't made any arithmetic errors in the wrong path 🙂

I think your first two answers are correct, then realized your question had nothing to do with them :-). You list the right formula. Really it should be
s = s0 + ut + 1/2at^2
where s0 is the initial displacement.
Normally, we assume that the starting point is s=s0=0 (For this question, you could have made the starting point s = 0, then ground is s = -1.25). It pops out of integrating a constant acceleration for the same reason we get
v - u = at or
v = u + at
u is the initial velocity. Integrate again, we get
s - s0 = ut + 1/2at^2 or
s = s0 + ut + 1/2at^2
Does this answer your question?

I think your first two answers are correct, then realized your question had nothing to do with them :-). You list the right formula. Really it should be
s = s0 + ut + 1/2at^2
where s0 is the initial displacement.
Normally, we assume that the starting point is s=s0=0 (For this question, you could have made the starting point s = 0, then ground is s = -1.25). It pops out of integrating a constant acceleration for the same reason we get
v - u = at or
v = u + at
u is the initial velocity. Integrate again, we get
s - s0 = ut + 1/2at^2 or
s = s0 + ut + 1/2at^2
Does this answer your question?

Why do you use u instead of v in the initial equation and please may you explain how you get from the first to the second equation you give via integration? 🙂

Integrating we get
v(t) - u = at
... Similarly when we integrate again as v(t) = ds/dt ....

Thanks! I get why my wrong path was wrong! 😄, the only thing that I don't understand now is, on the first line that I've quoted, which equation do you integrate to get to v - u = at? 🙂

Directly from the definition of acceleration. Acceleration is a constant "a", so
d^2s/dt^2 = dv/dt = a
Integrate once (with respect to t), you get
v(t) - v_0 = at
...

Note that really we assume that the calculation starts at t = 0, and if not, we should have
v(t) - v_0 = a(t - t_0)
where t_0 is the starting time and v(t_0) = v_0.
You'd get a similar extra term in the equation for s(t) ...

You're getting confused with differentiation 😂🙂, tbh once I realised that I understood it 😄

Don't think I am
https://physics.info/kinematics-calculus/

Then why did you use that differentiation fraction (I know it's not classed as a fraction) instead of the integration sign? 🙂 plus when you differentiate s₀ + ut + 0.5at^2 you get at + u 😄

Int(dv/dt,t) = v(t) - v_0
The fundamental theorem of calculus is that integration is the antiderivative.
You integrate dv/dt to get velocity.
Same going from v(t) to s(t).

Nathan hits a tennis ball straight up into the air from a height of 1.25 m above the ground.

The ball hits the ground after 2.5 seconds. Air resistance is neglected.

the first question is to I find the speed with which Nathan hits the ball which is 11.75 m per second.

The second question is to find the greatest height above the ground reached by the ball Which is 8.29m.

the third question is to then find the speed of the ball when it hits the ground. This is the question I'm having trouble with; I know the answer and the correct pathway to get to the answer, I just don't know why the method outlined below is incorrect - please help me with this 🙂

S = 0 is defined at ground level.

Correct path is with v = u + at

Wrong path is with s = vt - 0.5at², although the right formula to use is s = vt - 0.5at² + 1.25 because when t = 0, s = 1.25

I know I haven't made any arithmetic errors in the wrong path 🙂

What's v(t)?

see post 5.

TBH, its clearly explained in the document I linked in post 9. The notation is just about identical and the integration process is clearly explained. Why not read that carefully?

Sorry I didn't read that when you posted it, I just didn't think it was necessary at the time 🙂, although I don't think it explains my last question 🙂

Which question?
and why don't you think it explains it?

What v(t) means and there's no mention of it in the link 🙂

v is the velocity at time t so its a signal v(t)
u is the initial velocity, so its a constant. Could have called it v_0
v(0) = u
I left out the initial equation (acceleration is constant), i.e.
dv/dt = a
Integrating we get
v(t) - u = at
... Similarly when we integrate again as v(t) = ds/dt ....