Projectile from certain height

  • Thread starter gracy
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  • #1
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Formula for time of flight in projectile motion is 2usin theta/g but I think it is only applicable for when object is launched from ground i.e in the case below
Projectile-motion-animation.png

But when it is launched from certain height this formula is no longer in use.
images?q=tbn:ANd9GcRjK4KyeUxJLwX90VQcMPICChwZ8V-Z5iL49ZLZ4wSKeiaxe-sE6Q.jpg
 
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Answers and Replies

  • #2
ZapperZ
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Is there a question here?

Zz.
 
  • #3
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Am I right in my first post.
 
  • #4
nasu
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Yes, it does not work in the second case.
 
  • #6
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And what about range and maximum height.Formula for range=u^2sin 2 theta/g
and maximum height =u^2 sin^2 theta/2g
Are these formulas also not applicable for second case.
 
  • #7
nasu
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I think it would be more beneficial for you to try to answer these starting from the basic equations.
This is the way to study. Not memorizing some formulas for special cases.
 
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  • #8
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I think it would be more beneficial for you to try to answer these starting from the basic equations.
This is the way to study. Not memorizing some formulas for special cases.
I think ,you are right.I have done some calculations.Check whether I am correct.
 
  • #9
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I have divided the trajectory in to two parts.
upload_2015-4-17_12-20-43.png


Now
S sub y (distance covered in y direction)=V0 sin theta T +1/2 g T^2
h=V0 sin theta T+1/2 g T^2
where T =time of flight
Now we can solve for T as other variables are given.The above equation forms quadratic equation.so we will gwt two values for T one of which will be negative ,as time can never be negative we will take the other one(positive value of T)
(I have taken downward direction to be positive)
And S sub x=(distance covered in x direction)=V0 cos theta t
If we put t=T in the equation we get
Vo cos theta T
which is nothing but Range.
For maximum height
I have again divided the trajectory in two parts

upload_2015-4-17_12-37-0.png

Now we can see the maximum height=h+h'
where h' is same as in case of object projected from horizontal surface and lands on the same horizontal surface.And h is given,hence we can find maximum height.
Is my work fine?
 
  • #10
jbriggs444
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If you take the downward direction to be positive then your formula for Sy as a function of time has the wrong sign on the contribution from the object's initial velocity.
 
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  • #11
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If you take the downward direction to be positive then your formula for Sy as a function of time has the wrong sign on the contribution from the object's initial velocity.
It should be as follows.
h=V0 sin theta T+1/2 g T^2
h= - V0 sin theta T +1/2g T^2
Right?
 
  • #12
jbriggs444
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Right.
 
  • #13
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Thanks.
 

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