A stone dropped from a window reaches the ground in 0.5 sec.(g=10m/s²) .1) Calculate the speed just before it hits the ground.
2) What is the average speed during 0.5seconds?
3) Calculate the height of window from the ground
Answers
As per the given question ,
- Initial velocity = 0 m/s ( as the ball is just dropped )
- Time taken = 0.5 sec
- Acceleration due to gravity = 10 m/s ²
As the stone is moving uniformly under the influence of gravity we can use the kinematic equations in order to solve this question
(1) Speed of the stone just before it hit's the ground
By using the first kinematic equation ,
➽ v = u + at
here ,
- v = final velocity
- u = initial velocity
- a = acceleration
- t = time
➽ v = 0 + 10 x 0.5
➽ v = 5 m/s
The speed of the ball just before it hit's the ground is 5 m/s
2) Height of the window from the ground
By using the second kinematic equation ,
➸ h= ut + gt ²/2
here ,
- h = height
- u = initial velocity
- a = acceleration
- t = time
➸ h = 0 x 0.5 + 10 x 0.25 /2
➸ h = 5 x 0.25
➸ h = 1.25 m
The height of the window from the ground is 1.25 m
3) Average speed
As per the formula
➺ Average speed = total distance / total time
➺ Average speed = 1.25 / 0.5
➺ Average speed = 2.5 m/s
The average speed during 0.5 s is 2.5 m/s
A stone dropped from a window reaches the ground in 0.5 sec.(g=10m/s²) .
1) Calculate the speed just before it hits the ground.
2) What is the average speed during 0.5seconds?
3) Calculate the height of window from the ground.
Given:-
t = 0.5 sec ; a = 10m/s² ; u = 0
i. using first law of motion
➣ v = u + at
➣ v = 0 + 10 × 0.5
➣ v = 5m/s
ii. Average Speed
= distance travelled
time taken
Using second law of motion to compute distance travelled.
➣ s = ut + 1/2 at²
➣ s = 1/2 × 10 × 0.5 × 0.5
➣ s = 1.25m
Hence average Speed = 1.25/0.5 => 2.5m/s
iii. Height of stone would be equal to the distance travelled.
And hence, height of stone is 1.25m
✐ Hence Verified