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Projectile being launched from ground level

  • Thread starter AryRezvani
  • Start date
  • #1
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Homework Statement



A projectile is fired from ground level with an initial velocity of 50 m/s and an initial angle of 37°. Assuming g = 9.8 m/s^2, find:

a) the projectiles total time of flight.
b) the maximum height attained
c) the total horizontal distance traveled
d) the final horizontal and vertical velocities just before it hit the ground.


Homework Equations



okay, so beginning with a), I believe the necessary equation is h=vi2Sin2θi/2g


The Attempt at a Solution



Plug in a) into my calculator, and I get uhh 4436.721196. I'm not sure if i'm using the correct method.
 

Answers and Replies

  • #2
3,812
92
okay, so beginning with a), I believe the necessary equation is h=vi2Sin2θi/2g
You need to find the total time of flight in a), the equation you posted is for the max. height attained by the projectile.
 
  • #3
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You need to find the total time of flight in a), the equation you posted is for the max. height attained by the projectile.
would you utilize Vf=Vi+at?

So Vf = 0 (at the top of the trajectory Vf would be zero.
g = -9.8
Vi = 50

Plus in so you get -9.8t = -50
t = 5.102?
 
  • #4
3,812
92
would you utilize Vf=Vi+at?

So Vf = 0 (at the top of the trajectory Vf would be zero.
g = -9.8
Vi = 50

Plus in so you get -9.8t = -50
t = 5.102?
For the vertical motion, what's the initial velocity? You need to take the component of initial velocity in the vertical direction for Vi.
 
  • #5
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For the vertical motion, what's the initial velocity? You need to take the component of initial velocity in the vertical direction for Vi.
Hmm I got 50*Sin(37) which gave me 30.09075116.
 
  • #6
3,812
92
Hmm I got 50*Sin(37) which gave me 30.09075116.
You shouldn't get such a big answer. The component you took is correct. Check your calculations. Maybe, it will help if you show us your steps. Make sure you double the time you get when you plug the values in the equation Vf=Vi+at because you need to find the total time of flight. The projectile reaches the max. height and comes back again.
 
  • #7
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Okay, well in my textbook, it says the following formula is utilized to find Vyi:

Vyi=ViSin(θi)

I plugged in the numerical values into my calculator and got Vyi = 30.09075116.

Now, I plugged that answer into Vyf = Vyi + ayt

Vyf should equal zero because at the top of the trajectory the y component is zero.

I solved and got t = 3.09075116 seconds.

Edit; multiply the answer by 2 and you get 6.14096.... Look good?
 
  • #8
3,812
92
Okay, well in my textbook, it says the following formula is utilized to find Vyi:

Vyi=ViSin(θi)

I plugged in the numerical values into my calculator and got Vyi = 30.09075116.
Oops, sorry, i thought you meant time of flight when you wrote 30.09075116. :redface:

Now, I plugged that answer into Vyf = Vyi + ayt

Vyf should equal zero because at the top of the trajectory the y component is zero.

I solved and got t = 3.09075116 seconds.
Just double the time you calculated, for the reason i mentioned in my previous post. :wink:

EDIT: Just saw your edit, looks good.
 
  • #9
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Oops, sorry, i thought you meant time of flight when you wrote 30.09075116. :redface:


Just double the time you calculated, for the reason i mentioned in my previous post. :wink:

EDIT: Just saw your edit, looks good.
Thanks :)

If you could, could you just quickly check my work for the rest of them?

In regards to b), I used the equation I mentioned initially (Vi2Sin2θi/2g

Plugged in (50^2)*Sin(37)*Sin(37)/2*9.8 and got 4436.721196.

Look a little off?
 
  • #10
3,812
92
Thanks :)

If you could, could you just quickly check my work for the rest of them?

In regards to b), I used the equation I mentioned initially (Vi2Sin2θi/2g

Plugged in (50^2)*Sin(37)*Sin(37)/2*9.8 and got 4436.721196.

Look a little off?
Its too much off. :biggrin:
Check your calculations again. Or show the steps.
 
  • #11
67
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Its too much off. :biggrin:
Check your calculations again. Or show the steps.
Ah, yeah, incorrect calculations lol. I fixed it and got something around 46.196..

Okay, so part c) I found the Vxi to be 39.931775 or does this have something to do with range (R)?
 
  • #12
3,812
92
Okay, so part c) I found the Vxi to be 39.931775 or does this have something to do with range (R)?
Yes, you need to find the range.
 
  • #13
67
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c) Vxi = 39.9317755

Formula for R being: Vxi2ta

Ta being the halfway point of the trajectory so all in all this comes out to be the time we calculated earlier.

R = (39.9317755)(3.07048412)2

R = 245.2198204 meters

Look good?
 
  • #14
3,812
92
c) Vxi = 39.9317755

Formula for R being: Vxi2ta

Ta being the halfway point of the trajectory so all in all this comes out to be the time we calculated earlier.

R = (39.9317755)(3.07048412)2

R = 245.2198204 meters

Look good?
Looks correct to me.
 
  • #15
67
0
Looks correct to me.
Okay, lastly, d).

Hmm okay, well horizontal velocity is constant so I don't think there is a way of doing the horizontal one.

Wouldn't these just be their respective initial velocities? Or maybe you'd plug them into an equation that has a Vf and solve.

Vf = Vi + at

Time being how long the ball is in the air.
 
  • #16
3,812
92
Okay, lastly, d).

Hmm okay, well horizontal velocity is constant so I don't think there is a way of doing the horizontal one.

Wouldn't these just be their respective initial velocities? Or maybe you'd plug them into an equation that has a Vf and solve.

Vf = Vi + at

Time being how long the ball is in the air.
Horizontal velocity will be the same, vertical velocity will be opposite in direction to that of initial. It asks about velocity, not speed.
 
  • #17
67
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Horizontal velocity will be the same, vertical velocity will be opposite in direction to that of initial. It asks about velocity, not speed.
Understood, so the object fired has an initial velocity of +50 m/s; however, as it comes down, the velocity is the same, but simply decreasing?

So the velocity just as the object hits the ground is -50 m/s?

Edit: Leaving for class right now. Thanks for taking the time to explain this to me. I really appreciate the help. :)
 
  • #18
3,812
92
Understood, so the object fired has an initial velocity of +50 m/s; however, as it comes down, the velocity is the same, but simply decreasing?

So the velocity just as the object hits the ground is -50 m/s?
No, i did not mean that. The question asks about the vertical and horizontal final velocities. The direction of final vertical velocity would be the opposite to that of the initial vertical velocity. You cannot say that its -50 m/s.
 
  • #19
67
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No, i did not mean that. The question asks about the vertical and horizontal final velocities. The direction of final vertical velocity would be the opposite to that of the initial vertical velocity. You cannot say that its -50 m/s.
So what would it be? Would it be 50 m/s simply on the opposite direction? In other words, 50 m/s with a downward direction?
 

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