A coin is dropped from a window. If it hits the ground at 10 m/s, (i) work out the height of the window (i.e distance travelled). Take acceleration due to gravity as 10m/s2. (ii) Find also the time taken to reach the ground.
Answers
Given :
- Final Velocity of the coin = 10 m/s.
- Intial Velocity of the coin = 0 m/s.
- Accelaration of the coin = 10 m/s².
To find :
- Height of the window.
- Time Taken to reach the ground.
Solution :
To find the height of the window :
We know the Third Equation of Motion,i.e,
⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² ± 2gh
Where :
- v = Final Velocity.
- u = Initial Velocity.
- g = Acceleration due to gravity.
- h = Height.
Now by using the Third Equation of Motion and substituting the values in it, we get :
==> v² = u² ± 2gh
[Note : Here the Acceleration due to gravity will be taken as postive, since we the body is moving with the gravity]
==> v² = u² + 2gh
==> 10² = 0² + 2 × 10 × h
==> 100 = 0 + 20h
==> 100 = 20h
==> 100/20 = h
==> 5 = h
∴ h = 5 m.
Time Taken by the coin :
We know the second Equation of the motion .i.e,
⠀⠀⠀⠀⠀⠀⠀⠀⠀h = ut + ½gt²
Where :
- v = Final Velocity.
- t = Time taken
- g = Acceleration due to gravity.
- h = Height.
Now by using the Third Equation of Motion and substituting the values in it, we get :
==> h = ut + ½gt²
==> 5 = 0(t) + ½gt²
==> 5 = 0 + ½gt²
==> 5 = ½gt²
==> 5 = ½ × 10 × t²
==> 5 = 5 × t²
==> 5/5 = t²
==> 1 = t²
==> √1 = t
==> ± 1 = t
∴ t = 1 [Since , time taken cannot be negative]
Hence the time taken by the body to reach the ground is 1 s.