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?

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;

where;

g is acceleration due to gravity, which is constant

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

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

magnityde and direction

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)