Question

A ball falls off a table and reaches the ground in 1 s.Assuming g=10m/s^2, calculate it's speed on reaching the ground and the height of the table.

Answer 1

$v = u +at \\ v = gt \\ v = 10m {s}^{ - 1}$
$h = \frac{1}{2} g {t}^{2} \\ h = 5m$
You can verify by using:
${v}^{2} - {u}^{2} = 2gh \\ {v}^{2} = 100 \\ v = 10m {s}^{ - 1}$
In 2gh, we put the value of h=5m i.e. the height we got.

Answer 2

In answering this question we need to use kinematics equation.

We need two things ; The height traveled and the final velocity.

The initial velocity in this case equals to 0.

1) Velocity on reaching the ground

v = u + gt

v = final velocity

u = initial velocity

t = time

g = gravitational acceleration.

Doing the substitution we have :

v = 0 + 10 × 1 = 10 m/s

= 10 m/s

2) The height of the table

S = ut + 1/2gt²

Doing the substitution we have :

S = 0 × 1 + 1/2 × 10 × 1

s = 0 + 5

= 5 meters