Question

A ball released from rest from the roof of
a building of height 45m at t = 0. Find the
height of ball from ground at t = 2s (g =
10 m/sec)​

Answer 1

Given :

▪ A ball is released from rest from the roof of a building.

▪ Height of building = 45m

▪ Acc. due to gravity = 10m/s²

To Find :

▪ Height of ball from ground at t = 2s

Concept :

☞ Acceleration due to gravity acts continuously throughout the motion, we can easily apply equation of kinematics to solve this question.

☞ For a body falling freely under the action of gravity, g is taken positive.

✴ Second equation of kinematics :

$\bigstar\:\underline{\boxed{\bf{\red{s=ut+\dfrac{1}{2}at^2}}}}$

• s denotes distance
• u denotes initial velocity
• a denotes acceleration
• t denotes time

CalculaTion :

$\dashrightarrow\sf\:H=ut+\dfrac{1}{2}gt^2\\ \\ \dashrightarrow\sf\:H=(0\times 2)+\dfrac{1}{2}(10\times 2^2)\\ \\ \dashrightarrow\sf\:H=0+\dfrac{40}{2}\\ \\ \dashrightarrow\bf\:H=20m\\ \\ \dashrightarrow\sf\:Height\:from\:ground\: (H')=45-20\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{H'=25m}}}}\:\orange{\bigstar}$