# Right Triangle Question for physics problem

1. Oct 2, 2009

### 1irishman

1. The problem statement, all variables and given/known data
given only the time in the air and the range horizontally of one projectile. How can i find the angle and velocity of the projectile?

2. Relevant equations
trig functions?

3. The attempt at a solution
not sure how...hints?

2. Oct 2, 2009

### Delphi51

Just begin with angle A and initial velocity V and proceed boldly as if you know the numbers. Separate the initial velocity into horizontal and vertical parts. Then write two headings:
Horizontal and Vertical. Decide in each case whether you have constant speed or accelerated motion and write the appropriate
formula(s). Put in the numbers or expressions for all known
quantities and look for an equation you can solve.

3. Oct 2, 2009

### 1irishman

Hmmm...okay...well, i'm thinking i might be able to use one or both of these equations to start:
d=vit+1/2at^2
vf=vi+at
Now, i have drawn two headings labeled horizontal and vertical and this is where i find it really tricky:
under horizontal i have put these values: vi=0, a=0, d=50m, t=3
under vertical i have put these values: vi=0, a=-9.80m/s^2, vf=0, d=0,t=3

i'm confused as to which heading i should place distance and time under both or just one of them? Also not sure about my other values? More hints please?

4. Oct 2, 2009

### 1irishman

Sorry, the original problem gave a range of 50m and total air time of 3s.

5. Oct 2, 2009

### Delphi51

Put the numbers in all three formulas. Just d = vt for horizontal.
Looks like the horizontal formula will give you a numeric value for v*cos(A) and one of the vertical ones will give you v*sin(A). Knowing both initial velocity components you can quickly find the magnitude and angle.

6. Oct 2, 2009

### 1irishman

Do you mean put 3s and 50m in all three formulas? For horizontal velocity i got 16.7m/s.

7. Oct 2, 2009

### Delphi51

Yes, agree. Vertical velocity 14.7.

8. Oct 2, 2009

### 1irishman

I see, so the hypotenuse should have value 16.7m/s and the adjacent side should have value 50? Thanks.

9. Oct 2, 2009

### bleedblue1234

range of 50m and total air time of 3s.

So the vertical component is always independent of the horizontal component... so

Vx = Dx / t

so Vx = 50/3 m/s

so...

(delta) y = Vyot+1/2gt^2

0 = Vyo(3)+1/2(-9.80)9

0 = Vyo(3) -44.1

44.1/3 = Vyo

14.7 = Vy

so Vx = (50/3) m/s and Vy = (14.7) m/s

so Vr = sqrt(Vx^2 + Vy^2)

so Vr = 22.2 m/s

so for angle you can just Sin-1 of (14.7/(50/3))

so your angle of launch is 61.9 degrees.

10. Oct 2, 2009

### 1irishman

oh okay...hmm...the answer in the book says 41 degrees is angle of launch.

11. Oct 2, 2009

### 1irishman

i used sin-1 14.7/22.2=41 degrees

12. Oct 2, 2009

### Delphi51

I also get 41.4. Bleedblue should have used inverse tan instead of inverse sin in his last step.

13. Oct 2, 2009

### 1irishman

Thanks Delphi