Question

If an object free-falls for t seconds from rest to d distance, how far will the object fall from rest in twice the elapsed time?

Answer 1

Answer:

When the time of fall is doubled, the height of fall will be quadrupled

Explanation:

Given;

height of fall, h = d m

time of fall, t = t s

initial velocity of the object, u = 0 m/d

The height of fall of the object is calculated from the kinematic equation below;

$h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\g = \frac{2h}{t^2}$

where;

g is acceleration due to gravity, which is constant

if the time of fall is doubled, the height of fall is calculated as;

$\frac{2h_1}{t_1^2} = \frac{2h_2}{t_2^2} \\\\\frac{h_1}{t_1^2} = \frac{h_2}{t_2^2}\\\\(Note: h_1 = d, \ and \ t_1 = t)\\\\h_2 = \frac{h_1t_2^2}{t_1^2} \\\\h_2 = \frac{(d)(2t_1)^2}{t_1^2} \\\\h_2 = \frac{(d)(2t)^2}{t^2}\\\\h_2 = \frac{(d)\times 4t^2}{t^2}\\\\h_2 = 4d$

Therefore, when the time of fall is doubled, the height of fall will be quadrupled