# Angle of launch

1. Jul 2, 2011

### mcr

Hi,
I'm new at this and I need some help with the following practice problems:

1) At what angle relative to the horizontal must an object be
launched if its minimum velocity in flight is 28% of its launch
velocity? The answer is to be in degrees. I have the answer to be 73.7.

2) During a footrace, a fast runner runs at a 6 min/mi pace, and a
slow runner runs at a 7 min/mi pace. The fast runner passes the slow
runner. After a 2 sec delay, the slow runner initiates an acceleration,
catching up with the fast runner in 968 ft. What is this acceleration?
The answer is to be in ft/sec^2 and the answer is .0798.

I'm confused on both and don't know how to get started.

2. Jul 2, 2011

### tiny-tim

welcome to pf!

hi mcr! welcome to pf!
yup!
write two equations, one for each runner, both starting at the time that they are level

(so the second one will have constant speed for t ≤ 2, and https://www.physicsforums.com/library.php?do=view_item&itemid=204" for t > 2 )

Last edited by a moderator: Apr 26, 2017
3. Jul 2, 2011

### Andrew Mason

What is the condition for minimum total velocity? (hint: does the horizontal velocity change during flight - ignoring air resistance -?)

Convert the speeds into feet/sec. Draw a graph of speed vs. time for each runner starting at the moment the fast runner passes. What does the area under the graph represent?

Write out the expression for area under the graph for each runner. What is the condition (in terms of the areas under each of the graphs) for the slow runner to catch the fast one?

AM

4. Jul 2, 2011

### mcr

ok, I'm needing more assistance.

problem 1: I'm assuming the minimum total velocity is at the peak of its flight. Is this correct? But, I'm still stuck.

problem 2: for the fast runner I have d=14.67(ft/sec)t+4.2 and for the other runner I have d=1/2at^2. I'm not sure on these. That's all I have.

5. Jul 2, 2011

### Andrew Mason

Write out the expression for total velocity in terms of $\vec{v_y} \text{ and } \vec{v_x}$. (hint: you have to add the horizontal and vertical velocities as vectors). How does this total velocity change with time?

Where does the 4.2 come from?
Not quite. For t<2 what does the graph of velocity vs time look like for the second runner? What is the area under the graph for that part? What does the graph look like for t>=2? What is the area under the graph for that part? Add the areas together. When that is equal to the area under the first graph, what has occurred?

AM

6. Jul 3, 2011

### azureth

1:The minimum of its velocity is exactly when it is at the peak(you can prove it according to the law of conservation of mechanical energy).Here you may find Vx is V*cosa and Vy is zero,therefore Vmin is V*cosa(V stands for initial velocity).

7. Jul 3, 2011

### mcr

Thanks, I finally got #1. On to #2. The velocity vs time graph for one runner is going to be a horizontal line. I found it to be y=14.67. The velocity vs time graph for the other runner has two pieces. For t<=2 I have y=12.57, but for t>2 it is a slanted line (y=mx+b), but I don't know how to get the parameters. This is my stumbling part as of right now.

8. Jul 3, 2011

### Andrew Mason

What does the slope represent?

Let the slope be a. What is the area under that part of the graph in terms of a?

AM

9. Jul 3, 2011

### mcr

So the slope is the acceleration, but what about b?

10. Jul 3, 2011

### Andrew Mason

Right - slope is acceleration. Don't worry about b. You just need to work out the equation for area under the graph for t>2 (hint: it consists of a triangle and a rectangle - work out the area of each as a function of time t' = t-2)

AM

11. Jul 4, 2011

### mcr

Ok, this is what I got for the area under the curves which represents their distances. For the faster runner I got d=14.7t. For the slower runner I got d=12.6t+1/2a(t-2)^2. So now the two distances are equal so 14.7t=12.6t+1/2a(t-2)^2. Is this correct? I have one equation with two unknowns. So I need to solve for t by coming up with another equation. This is where I am lost. All help is appreciated.

12. Jul 4, 2011

### Andrew Mason

That looks right.

Have another look at the problem. What is the distance covered by the slow runner (after 2 seconds)?

It might be easier to break it down:

Distance travelled by slow runner in first two seconds = 12.6 x 2 = 25.2 feet
Distance travelled by slow runner after 2 seconds = 12.6(t-2) + .5a(t-2)^2

So the equation is: 14.7t = 25.2 + 12.6(t-2) + .5a(t-2)^2

AM

13. Jul 4, 2011

### mcr

So I proceeded to use the equation: 12.6(t-2)+.5a(t-2)^2=198 ft. and solved for (t-2) in terms of a. I then substituted back into the equation and I get a=.0663 ft/sec^2. This is not the answer, so I must have done something incorrect. Does anything stand out that is wrong? Thanks.

14. Jul 10, 2011

### Andrew Mason

?? Where does the 198 feet come from? Your statement says 968 feet. Your answer is correct if the question says 968 feet. Better check the question again.

AM

Last edited: Jul 10, 2011