02. A ball is dropped from a height of 60 m. Calculate i) the time taken by the ball to reach the
ground, and (ii) the velocity with which the ball strikes the ground.
Answers
GIVEN:
Height = 60m
Initial velocity = 0 m/s
Assuming 'a' = 10m/s²(acceleration due to gravity)
(i)
USING FORMULA:
h = ut + ½gt²
(Where: h=height, g=acceleration due to gravity)
60 = (0×t) + ½×10×t²
60 = 0 + 5t²
t² = 60÷5
t² = 12
t = (sqrt)12
(ii)
v = u + at
v = 0 + 10×(sqrt)12
v = 10 root 12
❮ Answer :❯
- Time taken by ball to reach ground = 3.5 seconds.
- Velocity with which ball strikes the ground = 34.3 m/s.
❮ Given :❯
- Height from which ball is dropped, s = 60 m
- Initial velocity of ball, u = 0
- acceleration acting on the ball = acceleration due to gravity = 9.8 m/s^2
❮ To find :❯
- Time taken by ball to reach the ground, t = ?
- Velocity with which ball strikes the ground, v = ?
❮ Knowledge required :❯
✪ Equations of motion ✪
- First equation of motion
v = u + a t
- Second equation of motion
s = u t + 1/2 a t^2
- Third equation of motion
2 a s = v^2 - u^2
⁃ [ where u is initial velocity, v is final velocity, s is distance covered, t is time taken and a is acceleration of the body ]
❮ Solution :❯
Using second equation of motion
➥ s = u t + 1/2 a t^2
➥ 60 = ( 0 ) t + 1/2 ( 9.8 ) t^2
➥ 60 = 4.9 t^2
➥ t^2 = 60/4.9
➥ t = (2√(15) / 0.7) sec
➥ t = 3.5 sec (approx.)
therefore,
- Time taken by ball to reach the ground is 3.5 seconds approximately.
Now,
Using third equation of motion
➥ 2 a s = v^2 - u^2
➥ 2 ( 9.8 ) ( 60 ) = v^2 - ( 0 )^2
➥ 1176 = v^2
➥ v = 14 √6 m/s
➥ v = 34.3 m/s (approx.)
therefore,
- velocity with which ball strikes the ground will bd 34.3 m/s approximately.