Projectile from certain height

In summary, the formula for time of flight in projectile motion is 2usin theta/g, but it is only applicable when the object is launched from the ground. When launched from a certain height, this formula is no longer valid. The formulas for range and maximum height in projectile motion are also not applicable in this case. It is more beneficial to understand and solve the equations from the basic principles, rather than memorizing formulas for specific scenarios. To calculate the time of flight in this case, the trajectory can be divided into two parts and the quadratic equation can be solved for T. The formula for maximum height is h=V0 sin theta T+1/2 g T^2, taking the downward direction as positive. When solving
  • #1
gracy
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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
 
Last edited:
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  • #2
Is there a question here?

Zz.
 
  • #3
Am I right in my first post.
 
  • #4
Yes, it does not work in the second case.
 
  • #5
  • #6
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
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
nasu said:
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
I have divided the trajectory into 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
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
jbriggs444 said:
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.
gracy said:
h=V0 sin theta T+1/2 g T^2
h= - V0 sin theta T +1/2g T^2
Right?
 
  • #12
Right.
 
  • #13
Thanks.
 

Related to Projectile from certain height

What is a projectile?

A projectile is any object that is launched into the air and moves along a curved path under the influence of gravity.

What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by the initial velocity, angle of launch, and the force of gravity.

How is the height of a projectile calculated?

The height of a projectile can be calculated using the equation h = h0 + v0t + 1/2gt2, where h0 is the initial height, v0 is the initial velocity, t is time, and g is the acceleration due to gravity.

What is the maximum height of a projectile?

The maximum height of a projectile occurs at the highest point of its trajectory, where the vertical velocity is zero. This can be calculated using the equation hmax = v02/2g.

How does air resistance affect the trajectory of a projectile?

Air resistance can slow down a projectile and cause it to travel a shorter distance. It can also change the shape of the trajectory, making it less predictable.

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