You’re a consultant on a movie set, and the producer wants a car
to drop so that it crosses the camera’s field of view in time
The field of view has height h. Derive an expression for the
height above the top of the field of view from which the car
should be released.

help me here please

Question

You’re a consultant on a movie set, and the producer wants a car
to drop so that it crosses the camera’s field of view in time
The field of view has height h. Derive an expression for the
height above the top of the field of view from which the car
should be released.

help me here please

Ultrilliam · Answer

@sweetburger Can you help this person?

sweetburger · Answer

Well wait a second I guess some assumptions can be made

bananachocolate · Answer

please try making assumptions then:P
im stuck since last night on this very question:P

sweetburger · Answer

Well what we can say is we definitely know we have to drop the car from some initial position of the car being at some higher point then the top of the field of view of the camera. So we can say here $y_o>y_f$. The difference between the initial height and the height of the top of the field of the view of the camera can be denoted as $ x_f-x_o=\Delta x $ (this will be important in just a moment). We also know that if we drop the car it has an initial velocity of zero, $v_o=0$ but we have an undefined $v_f$. A car ignoring air resistance is only subject to the action at a distance force gravity and therefore has a constant acceleration of $a_y=-9.8=-g$ 
Now we're going to arbitrarily say that the field of view of the camera has some height H. We want our $ x_f =H $. So the total distance we need to move is simply$\Delta x$ which represents the distance between the start of where the car was dropped and the top of the field of view of the camera. So using some simple kinematics equations we can plug in some of the knowns we discovered
$$x_f=x_io+ v_ot+\frac{ 1 }{ 2 }at^2$$
$$-\Delta x= v_ot+\frac{1}{2}(-g)t^2$$   isolating for $v_o$
$$v_o= \frac{ 2g t^2-2\Delta x }{ t }$$
now letting $$v_f=v_o$$
$$v_f^2=v_o^2+2a\Delta x$$
$$(\frac{ 2g t^2-2 \Delta x }{ t })^2=0+2(-g) (-\Delta x)$$
$$\Delta x =\frac{ (\frac{ 2g t^2 - 2 \Delta x }{ t })^2 }{ 2g }$$

sweetburger · Answer

going to go pass out

sweetburger · Answer

oh and uhh let x=y i guess i didn't keep that consistent whoops

sweetburger · Answer

no difference in meaning there

Ultrilliam · Answer

@bananachocolate

bananachocolate · Answer

OMG! you just saved a life;)
i finallly understand the question. thankyou soooo much!@sweetburger :P (pretty cool name btw:P)

sweetburger · Answer

haha no problem glad it helped