Mechanics 1 problems :/ Help PLEASE!

I got set these two questions based on modelling using constant acceleration.

I've spent a few hours today on these and I still can't arrive at the correct answer.

1) Nathan hits a tennis ball straight up into the air from a height of 1.25m above the ground. The ball hits the ground after 2.5seconds. Assuming g = 10ms-2, find

i) the speed Nathan hits the ball
ii) the greatest height above the ground reached by the ball
iii) the speed the ball hits the ground
iv) how high the ball bounces if it loses 0.2 of its speed on hitting the ground.


And

2)Towards the end of a half-marathon Sabina is 100m from the finish line and is running at a constant speed of 5ms-1. Daniel, who is 140m from the finish and is running at 4ms-1, decides to accelerate to try to beat Sabina. If he accelerates uniformly at 0.25ms-1 does he succeed?

For number 2, find how long it takes Sabina to finish (time=distance/speed) and then find how long it takes Daniel to finish (using one of the constant acceleration (SUVAT) equations. Have you tried this, or is there any particular difficulty you're having using this method?

And now they dont have to learn they can just copy answers....

2)Towards the end of a half-marathon Sabina is 100m from the finish line and is running at a constant speed of 5ms-1. Daniel, who is 140m from the finish and is running at 4ms-1, decides to accelerate to try to beat Sabina. If he accelerates uniformly at 0.25ms-1 does he succeed?

Question 1:

s=-1.25
u=?
v=?
a=10
t=2.5
(units omitted because I keep thinking the "s" in "m/s" is related to displacement)
(Positive direction is upwards, hence negative acceleration)

i) The initial speed upwards

s=ut+0.5at^2
-1.25=2.5u-31.25
2.5u=30
u=12

ii) The greatest height reached

Here, we find the position where the ball reached zero velocity due to gravity.

From the previous question, we can see that u=12, and in this case, v=0, and a=-10

v=u+at
0=12-10t
10t=12
t=1.2

Now, we find the distance travelled:

s=(u+v)t/2
s=12*1.2/2
s=7.2

iii) The speed of the ball upon first impact with the ground

We know at the top of the flight, velocity is zero, so set u=0.
We also know that a=10 (this time, positive velocity is downwards).

The total time is 2.5, and 1.2 of that is spent travelling upwards. This means that the ball fell 1.3 seconds.

v=u+at
v=10*1.3
v=13

iv) The greatest height after the first bounce

The ball bounces up at the speed of 12.8. We know at the top of its bounce, v=0, and we also know that a=-10

So:

v=u+at
0=12.8-10t
10t=12.8
t=1.28

And with s=t(u+v)/2:

s=1.28(12.8)/2
s=8.192

Hope that helps.

Question 1:

s=-1.25
u=?
v=?
a=10
t=2.5
(units omitted because I keep thinking the "s" in "m/s" is related to displacement)
(Positive direction is upwards, hence negative acceleration)

i) The initial speed upwards

s=ut+0.5at^2
-1.25=2.5u-31.25
2.5u=30
u=12

ii) The greatest height reached

Here, we find the position where the ball reached zero velocity due to gravity.

From the previous question, we can see that u=12, and in this case, v=0, and a=-10

v=u+at
0=12-10t
10t=12
t=1.2

Now, we find the distance travelled:

s=(u+v)t/2
s=12*1.2/2
s=7.2

Don't forget to add 1.25, as this was the starting point. So, the ball actually reaches 8.45m into the air.

iii) The speed of the ball upon first impact with the ground

We know at the top of the flight, velocity is zero, so set u=0.
We also know that a=10 (this time, positive velocity is downwards).

The total time is 2.5, and 1.2 of that is spent travelling upwards. This means that the ball fell 1.3 seconds.

v=u+at
v=10*1.3
v=13

iv) The greatest height after the first bounce

The ball bounces up at the speed of 12.8. We know at the top of its bounce, v=0, and we also know that a=-10

So:

v=u+at
0=12.8-10t
10t=12.8
t=1.28

And with s=t(u+v)/2:

s=1.28(12.8)/2
s=8.192

Hope that helps.

Question 2

i) Finding Sabina's time

s=100
u=5
a=0
t=?

s=ut+0.5at^2
100=5t
t=20

ii) Finding Daniel's time

s=140
u=4
a=0.25
t=?

s=ut+0.5at^2
140=4t+0.125t^2

Rearrange into a quadratic:

0.125t^2+4t-140=0

t cannot be negative, so using the quadratic formula, I got t=21.09

Daniel takes 1.09 seconds more than Sabina to cross the finish line, so he fails to win.

That's what I got. But the answer says: No, because he is 10 metres behind when sabina finishes

...

There are two ways you can look at the second question. Having worked out how long Sabrina takes to reach the finish line, you can either:

Work out how long Daniel takes to complete the rest of the course and compare the two, which is what you and Michi Kobayashi have done, or

Work out how far Daniel has run in the time Sabrina took to finish, and compare the distance he travelled with how far he needed to travel to reach the finish line, and that is what the book has done.

Either method is valid. The book's is simpler to work out.



Please see the guidelines on posting on the forum, at the top of the forum page, particularly in regard to posting full solutions straight off; it is against forum rules, with good reason.

THANK YOU
How did you know to do that?

I'm sorry for dragging this thread up again, but...

For finding the speed Nathan hits the ball, why can't you use v=u+at? It gives the wrong answer of 25 m/s but I don't see why it can't be used?

What values are you plugging into the equation? At the start, neither v nor u is known.