Question

a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time it takes for the container to reach the ground

Answer 1

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

$S=ut + \frac{1}{2}at^2$

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

$S=\frac{1}{2}gt^2$

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s$