NCERT Solutions for Class 9 Science Chapter 8
Motion
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Exercises Questions
Page No-113
Questions No-7
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Share
Here, u = 0 m/s, s = 20 m, a = 10 ms⁻², v = ?, t = ?
Using v² – u² = 2as
We have, v² – 0² = 2 × 10 × 20 = 400 ⇒ v = 20 ms⁻¹.
and t = (v – u) ÷ a = 20 ÷ 10 = 2 s.
For more answers visit to website:
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-8/
Given:
– Initial velocity u = 0 m/s (the ball is dropped, so initial velocity is 0)
– Acceleration a = 10 m/s²
– Distance s = 20 m (height from which the ball is dropped)
Final Velocity (when the ball strikes the ground):
We can use the equation of motion to find the final velocity (v)
v2 = u2 + 2as
v2 = 0 + 2 x 10 x 20
v2 = 400
v = √400
v = 20 m/s
Hence, the ball will strike the ground with a velocity of 20 m/s.
Time taken to strike the ground:
We can use another equation of motion to find the time (t) it takes for the ball to reach the ground.
v = u + at
20 = 0 + 10 x t
t = 20/10
t = 2 s
Therefore, the ball will strike the ground after 2 seconds.