# Thread: My answer multiplied by 2 is correct...what did I do wrong?

1. ## My answer multiplied by 2 is correct...what did I do wrong?

Question:
"A ball is thrown at 14.1 m/s at 45deg above the horizontal. Someone located 30m away along line of the path starts to run just as the ball is thown. How fast, and in which direction, must the person run to catch the ball at the level from which it was thrown?"

4.77m/s toward thrower

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I get -9.52 m/s which is double and I don't see any mistakes when I double things (I had to assumed symmetry and therefore multiplied by two for when Vfy was 0 so that I can get the full time interval for the horizontal component.

(My work is attached)

Any help would be greatly appreciated!

Edit: I meant my answer DIVIDED by 2 is correct (not multiplied).

2. Originally Posted by s3a
Question:
"A ball is thrown at 14.1 m/s at 45deg above the horizontal. Someone located 30m away along line of the path starts to run just as the ball is thown. How fast, and in which direction, must the person run to catch the ball at the level from which it was thrown?"

4.77m/s toward thrower

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

I get -9.52 m/s which is double and I don't see any mistakes when I double things (I had to assumed symmetry and therefore multiplied by two for when Vfy was 0 so that I can get the full time interval for the horizontal component.

(My work is attached)

Any help would be greatly appreciated!

Edit: I meant my answer DIVIDED by 2 is correct (not multiplied).

consider vertical motion , $\displaystyle s=ut+\frac{1}{2}at^2$

$\displaystyle 0=(u\sin \theta)t-\frac{1}{2}gt^2$

since $\displaystyle t=\neq 0$ , $\displaystyle t=\frac{2u\sin \theta}{g}=\frac{2(14.1\sin 45)}{9.81}$

=2.03 s

so this guy who has to run a distance of 9.7 m beginning from rest in 2.03 s with speed , v

$\displaystyle s=(\frac{v+u}{2})t$

$\displaystyle 9.7=\frac{v}{2}(2.03)$

v=9.56 m/s

3. I think that the book's answer is correct, because it (quite reasonably) assumes that the runner moves at a constant speed. To cover 9.7m in 2.03s you need to move at a speed of $\displaystyle 9.7/2.03 \approx 4.78$m/s.

4. So the whole problem is that I need to assume that the runner runs at a constant speed?

5. It would make most sense, yes.

"How fast" would be "at what speed".

The runner needn't start from zero m/s and
increase speed at a constant acceleration.
The ball is "subject" to acceleration, but the runner is on the ground,
and can choose speed.

If that or similar (changing speed in anticipation of the ball's position) was the case,
you could calculate an average speed.

6. Originally Posted by s3a
So the whole problem is that I need to assume that the runner runs at a constant speed?
yeah , my assumption is wrong , it's more reasonable to assume that the runner is running at a constant speed.

7. Do people run at an approximately constant speed in real life though? (under normal circumstances)

8. Originally Posted by s3a
Do people run at an approximately constant speed in real life though? (under normal circumstances)
no s3a,

it's a similar question to a car taking a certain time to cover
a particular distance.

The actual speed at any time depends on the circumstances
of the road, traffic, driver decisions, traffic lights etc.

Average speed is the speed at which the car would be travelling
if it was travelling at a constant speed the entire time.
This is an overall approximation based on the small amount
of information available.