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 t. 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.

Express your answer in terms of the variables h, t, and appropriate constants.

Answer 1

Let the velocity at the top of the field view be u.

acceleration = g
height = h

h = u * t + (1 / 2) * g * t²

let the height of release from the top of the field view be h_t
and the time from the release till reaching the top of the field view be t_t

from release,
u = g * t_t
h_t = (1 / 2) * g * t_t²

eleminating t_t we get,
u = sqrt (2 * g * h_t)

substituting values,
we get,
h = ((sqrt (2 * g * h_t)) + g * t ) * t - (1 / 2) * g * t²

solving for h_t we get,
h_t = [( 2h - gt² )² ] / [ 8gt² ]