# Homework Help: Problem with finding time from projectile motion problem

1. Jul 26, 2012

1. The problem statement, all variables and given/known data
A projectile is fired from a cliff 125m off the ground at 37° from the horizontal at 65 m/s. How long will it take to hit the ground?

2. Relevant equations
y= yo+(Vyo)(t) - 1/2(g)(t^2)

3. The attempt at a solution
substituting in equation....

(-125m)= 0 + (65m/s)(t) - 1/2 (9.8 m/s^2)(t^2)

(4.9 m/s^2)(t^2)- 65 m/s (t) -125m =0

rewrite in quadratic to solve for t

t=((65 m/s) +/- √((65)^2-4(4.9 m/s^2)(125)))/ 2(4.9m.s^2)

t = 10.9 s

answer in back of book is 10.4 s

not sure if using proper equation or totally missing something.

2. Jul 26, 2012

### tiny-tim

welcome to pf!

erm

where's the 37° ?

3. Jul 26, 2012

Heyas Tim,

The 37° is the angle the projectile is fired at from the top of the cliff with respect to the horizontal.

4. Jul 26, 2012

### tiny-tim

yes i know!

but it's not in your equations!!

5. Jul 26, 2012

So would I go for something like this...

To find the Vyo = (65 m/s)Sin 37 and Vxo = (65 m/s)Cos 37 instead of just using 65 m/s?

so then...
(-125m)= 0 + [(65m/s)sin 37](t) - 1/2 (9.8 m/s^2)(t^2) for the y component of time

and
(-125m)= 0 + [(65m/s)cos 37](t) - 1/2 (9.8 m/s^2)(t^2) for the x component of time?

Then do I just add them after using the quadratic for each?

6. Jul 26, 2012

### tiny-tim

yes
yes
erm

what is the acceleration in the x direction?
hmm … draw a diagram, and then think

what do the x and y equations each tell you?​

7. Jul 26, 2012

So should my acceleration be (+) and I would end up with an equation....
(-125m)= 0 + [(65m/s)cos 37](t) +1/2 (9.8 m/s^2)(t^2)

I drew a diagram and am at a loss for what each equation "should" be telling me.

8. Jul 26, 2012

### Saitama

As tiny-tim already said, what's the acceleration in x direction? g acts vertically downwards, is their any component of g in the x direction?

Show us the diagram you have drawn.

9. Jul 26, 2012

### swill777

This is wrong. The time it takes for the projectile to land depends only in the y direction. Therefore, your equation should reflect only the motion of the projectile in the y direction. your velocity should therefore be 65sin(37°)m/s instead of simply 65m/s as you have here. This is the mistake which accounts for the error in your answer.
I worked it out using 65sin(37°)m/s as the velocity and got 10.38s for the time.

10. Jul 26, 2012

I'll have to recheck my math. Thank you for the correction swill777!

11. Jul 27, 2012

Nope.... banging my head against the wall here... I'm missing something with my math.

12. Jul 27, 2012

HALLELUJAH!!!!

Nevermind now people. Go back to drinking coffee as I needed a whole ton to realize.....

http://roflrazzi.files.wordpress.com/2010/04/129164952429291808.jpg [Broken]

Thank you for your patience and help!

Last edited by a moderator: May 6, 2017
13. Jul 27, 2012

### PeterO

It is always a good idea to enter sin(30) into your calculator before each set of calculations just to check it gives you 0.5 - indicating if it is actually set to degrees.

14. Jul 27, 2012

### swill777

Yup, the degree/radian mix up will get ya. I've had to make it a habit to check my setting every time I deal with trig. Glad you got the right answer, though!