Question 103: Motion in 1D

Question

From 18 m height above the ground a ball is dropped from rest . The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is $\_\_\_\_$ m.
(Take $g=10 \mathrm{~m} / \mathrm{s}^2$ and neglect the air resistance)

Accepted Answer

Let the ball fall through a distance $s$ from the point of release.
Since it is dropped from rest,
initial velocity $u=0$
acceleration $a=g=10\ 	ext{m/s}^2$
We need the point where the magnitude of velocity becomes equal to the magnitude of acceleration.
So,
$$ v=g=10 $$
Now use the equation of motion:
$$ v^2=u^2+2as $$
Substituting the values:
$$ 10^2=0+2(10)s $$
$$ 100=20s $$
$$ s=5\ 	ext{m} $$
So the ball has fallen $5$ m from the top.
Initial height above ground $=18$ m
Hence height above ground at that instant:
$$ 18-5=13\ 	ext{m} $$
Therefore, the required height above the ground is
$$ \boxed{13\ 	ext{m}} $$

Motion in 1D

From 18 m height above the ground a ball is dropped from rest . The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is \_\_\_\_ m. (Take g=10 \mathrm{~m} / \mathrm{s}^2 and neglect the air resistance)

Integer / Numeric Answer Type