A particle is dropped from the top of a tower of height 4.9 m the velocity of particle with which it strikes the ground is
Share
A particle is dropped from the top of a tower of height 4.9 m the velocity of particle with which it strikes the ground is
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
We will use the following equation of motion.
V² = U² + 2gs
In this case :
V = Final velocity
U = initial velocity
g = gravitational acceleration
S = height of the tower.
The initial velocity equals to 0
Doing the substitution we have :
V² = 0 + 2 × 9.8 × 4.9 = 96.04
V² = 96.04
V = 9.8
= 9.8m/s
Given height of the tower h = 4.9 m.
Initial velocity u = 0.
g = 9.8 m/s^2.
We know that v^2 = u^2 + 2gh
= > v^2 = 2 * 9.8 * 4.9
= > v^2 = 96.04
= > v = 9.8.
Therefore, The velocity of the particle = 9.8 m/s.
Hope this helps!