Solve SUVAT Inconsistency: Find Time & Velocity of Ball

[mcairtime]

Hi there,

Sorry in advance. This problem is very simple but I can't see where I'm going wrong or what I'm missing. It may even be a numerical mistake, although I have run the numbers a few times. I have a feeling that I'm going to kick myself but anyway...

A ball is projected vertically downwards at 4 m/s from the top of a tower 80 m high. Find a) the velocity of the ball as it hits the ground and b) the time it takes to reach the ground.

(Taking down to be the positive direction and using g = 9.8)

a) use v^2 = u^2 +2as to find v = 12*sqrt(11) m/s or roughly 39.8 m/s (having discarded the negative solution).

b) use t = (v-u)/a to find t = (60*sqrt(11) - 20)/49 or approx 3.65 s

All fine, all easy.

Now, doing the problem again but using s = ut +0.5at^2 I get two solutions for t, one positive (matching the solution above) and a negative solution which may be discarded. All fine still.

When I try using s = vt - 0.5at^2 I get two solutions namely t = (60*sqrt(11) - 20)/49 i.e. matching the solution above but my second solution is t = (60*sqrt(11) + 20)/49 or roughly 4.47 s.

Why am I coming out with a second positive solution? I'll put my working below in case it's an arithmetic error:

s = vt - 0.5at^2

80 = 12*sqrt(11)t - (0.5)(9.8)t^2

4.9t^2 - 12*sqrt(11)t +80 = 0

t = (12*sqrt(11) +/- sqrt((12*sqrt(11))^2 - (4)(4.9)(80)))/9.8

t = (12*sqrt(11) +/- sqrt(16))/9.8

t = (60*sqrt(11) +/- 20)/49

Sorry about this.

 

[Doc Al]

s = vt - 0.5at^2

80 = 12*sqrt(11)t - (0.5)(9.8)t^2

Not quite sure what you are doing here. Where's your starting point? You've changed the sign of the acceleration, so up is positive.

 

[mcairtime]

Hi Doc Al,

Thanks for your response. I'm afraid I am not quite following. As far as I'm aware I'm maintaining down as the positive direction throughout the entire question.

In part a) we have

s = +80 m
u = +4 m/s
v = ?
a = +9.8 m/s^2
t = ?

We use this to find that v = + 12sqrt(11) m/s

For part b) we then have

s = +80 m
u = +4 m/s
v = +12sqrt(11) m/s
a = +9.8 m/s^2
t = ?

and I am getting a funny solution when I plug this into

s = vt - 0.5at^2

The minus sign is part of the equation, but I am subbing in a = + 9.8 m/s^2.

Apologies again if I'm being extra stupid.

 

[haruspex]

consider four points in a broader trajectory.
At A, it is at ground, rising at 39.8 m/s. At B, it has risen to 80m, rising still at 4m/s. C and D are the start and end points of the original problem.
Your solution using u gave the times for the C to D and C to A transitions (the second being negative). The s, u and a inputs are the same for both.
Your v equation solution gives the times for the transitions C to D and B to D. The s, v and a inputs are the same for both.

 

[Doc Al]

In part a) we have

s = +80 m
u = +4 m/s
v = ?
a = +9.8 m/s^2
t = ?

We use this to find that v = + 12sqrt(11) m/s

For part b) we then have

s = +80 m
u = +4 m/s
v = +12sqrt(11) m/s
a = +9.8 m/s^2
t = ?

All this is fine. You've solved for the final speed and the time.

and I am getting a funny solution when I plug this into

s = vt - 0.5at^2

The minus sign is part of the equation, but I am subbing in a = + 9.8 m/s^2.

I don't understand what you mean when you say "The minus sign is part of the equation". The basic kinematic equation is y = y₀ + v₀t + 0.5at^2. If you use down as positive, then there will be no minus sign.

Please explain what you are solving for at this point.

 

[mcairtime]

consider four points in a broader trajectory.
At A, it is at ground, rising at 39.8 m/s. At B, it has risen to 80m, rising still at 4m/s. C and D are the start and end points of the original problem.
Your solution using u gave the times for the C to D and C to A transitions (the second being negative). The s, u and a inputs are the same for both.
Your v equation solution gives the times for the transitions C to D and B to D. The s, v and a inputs are the same for both.

Yes, that is all crystal clear now. I was having a complete mental block. Seems obvious now. Thanks very much indeed.

 

[mcairtime]

All this is fine. You've solved for the final speed and the time.I don't understand what you mean when you say "The minus sign is part of the equation". The basic kinematic equation is y = y₀ + v₀t + 0.5at^2. If you use down as positive, then there will be no minus sign.

Please explain what you are solving for at this point.

Hello,

I have worked out where my issue was with the help of haruspex. As well as the equation s = ut + 0.5at^2 there is another equation s = vt - 0.5at^2 (where u is initial velocity and v is the final velocity). I was getting different answers when I used these two different equations and couldn't work out why. Apologies if I was unclear and thanks again.