Problem with finding time from projectile motion problem

[coneheadceo]

Homework Statement

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?

Homework Equations

y= yo+(Vyo)(t) - 1/2(g)(t^2)
quadratic formula

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.

 

[tiny-tim]

welcome to pf!

hi coneheadceo! welcome to pf!

erm …

where's the 37° ? ​

 

[coneheadceo]

Heyas Tim,

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

 

[tiny-tim]

yes i know!

but it's not in your equations! ​

 

[coneheadceo]

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?

 

[tiny-tim]

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?

yes

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

yes

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

erm …

what is the acceleration in the x direction? ​

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

hmm … draw a diagram, and then think …

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

 

[coneheadceo]

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.

 

[Saitama]

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.

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.

 

[swill777]

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

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.

 

[coneheadceo]

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

 

[coneheadceo]

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

 

[coneheadceo]

HALLELUJAH!


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


my calculator was set for answers in RAD instead of DEG

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

Thank you for your patience and help!

 

[PeterO]

HALLELUJAH!

my calculator was set for answers in RAD instead of DEG

Thank you for your patience and help!

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.

 

[swill777]

HALLELUJAH!


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


my calculator was set for answers in RAD instead of DEG

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!