Q. A ball is dropped from a height of 100 M. How much time will it take to reach the ground? Find the velocity of ball just before hitting the ground.
To find the time it takes for the ball to reach the ground and the velocity of the ball just before hitting the ground, we can use the equations of motion under constant acceleration due to gravity.
Given:
Initial height (h) = 100 meters
Acceleration due to gravity (g) = 9.8 m/s^2 (approximate value on the Earth's surface)
Time taken to reach the ground (t):
We can use the equation of motion: h = (1/2)gt^2
Where:
h = Initial height (100 meters)
g = Acceleration due to gravity (9.8 m/s^2)
t = Time taken to reach the ground (unknown)
Rearranging the equation to solve for t:
t^2 = (2h) / g
t^2 = (2 * 100) / 9.8
t^2 = 20.40816327
t ≈ √20.40816327
t ≈ 4.52 seconds (approximate value)
So, it will take approximately 4.52 seconds for the ball to reach the ground.
Velocity of the ball just before hitting the ground (v):
We can use the equation of motion: v = gt
Where:
v = Final velocity just before hitting the ground (unknown)
g = Acceleration due to gravity (9.8 m/s^2)
t = Time taken to reach the ground (4.52 seconds)
Substituting the values:
v = 9.8 m/s^2 * 4.52 s
v ≈ 44.496 m/s (approximate value)
The velocity of the ball just before hitting the ground is approximately 44.496 meters per second.
Answers & Comments
Verified answer
Answer:
44.496 meters per second
Explanation:
To find the time it takes for the ball to reach the ground and the velocity of the ball just before hitting the ground, we can use the equations of motion under constant acceleration due to gravity.
Given:
Initial height (h) = 100 meters
Acceleration due to gravity (g) = 9.8 m/s^2 (approximate value on the Earth's surface)
Time taken to reach the ground (t):
We can use the equation of motion: h = (1/2)gt^2
Where:
h = Initial height (100 meters)
g = Acceleration due to gravity (9.8 m/s^2)
t = Time taken to reach the ground (unknown)
Rearranging the equation to solve for t:
t^2 = (2h) / g
t^2 = (2 * 100) / 9.8
t^2 = 20.40816327
t ≈ √20.40816327
t ≈ 4.52 seconds (approximate value)
So, it will take approximately 4.52 seconds for the ball to reach the ground.
Velocity of the ball just before hitting the ground (v):
We can use the equation of motion: v = gt
Where:
v = Final velocity just before hitting the ground (unknown)
g = Acceleration due to gravity (9.8 m/s^2)
t = Time taken to reach the ground (4.52 seconds)
Substituting the values:
v = 9.8 m/s^2 * 4.52 s
v ≈ 44.496 m/s (approximate value)
The velocity of the ball just before hitting the ground is approximately 44.496 meters per second.
It is given that a ball is dropped from a height of 100m.
We need to find the time it will take to reach the ground as well as the velocity of the ball just before hitting the ground.
Solution :
Distance (s) = Height = 100 m.
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) = 0 m/s
Using the second equation of motion, we get :
s = ut + 1/2 (at²)
⇢ 100 = 0(t) + 1/2(10t²)
⇢ 100 = 1/2(10t²)
⇢ t² = 100(2)/10
⇢ t² = 20
⇢ t = √20
⇢ t = 4.47 seconds
Hence, the time taken to reach the ground is 4.47 sec (approx).
Now,
Using the first equation of motion :
v = u + at
⇢ v = 0 + 10(4.47)
⇢ v = 44.7 m/s
Hence, the velocity of the ball just before hitting the ground is 44.7 m/s (approx).
Important Note :
Here, equations of motion are taken when acceleration (a) is equal to the constant acceleration due to gravity (g).
This answer is calculated when g is taken as 10 m/s² and not 9.8 m/s².