Question

13 A ball drops from a building to the ground in 2 seconds. Let g = 10m , then
What is its speed on striking the ground
What is iis average speed for 2 seconds
How high is the building from the ground​

Answer 1

Answer:

It's speed on striking ground = 20 m/s

average speed for 2s = 10 m/s

Height of building = 20 m

Step by Step Explanations:

Given that,

A ball drops from a building to the ground in 2 seconds

here,

initial velocity of the ball = 0 m/s

let the final velocity of the by hitting ground be v

time taken to hit ground = 2s

here gravitational acceleration

= 10 m/s²

now we have,

Initial velocity(u) = 0

final velocity(v) = v

gravitational acceleration(g) = 10 m/s²

by the gravitational equation of motion

v = u + gt

putting the values,

v = 0 + 10(2)

v = 20 m/s

________________

now,

let the height of building be h

also,

v² = u² + 2gh

again putting the values

20² = 0² + 2(10)h

20h = 400

h = 400/20

h = 20 m

_________________

Now,

Average speed = (v + u)/2

= (20 + 0)/2

= 10 m/s

___________________

It's speed on striking ground = 20 m/s

____________________________

average speed for 2s = 10 m/s

_________________________

Height of building = 20 m

___________________

Answer 2

Answer:-

Given:-

Ball is dropped.so initial velocity is zero.

$u = 0$

$g = 10 \frac{m}{ {s}^{2} }$

Explanation:-

Case-1.

Final velocity = ?.

Now, From first kinematics equation we get,

$v = u + at$

Ball is coming down under influence of gravity so acceleration is equal to acceleration due to gravity.

$v = gt$

$v = 10 \times 2 = 20$

so, ball strikes the ground with 20m/s.

$\boxed{\boxed{v = 20m/s}}$

Case-2.

Height of building = ?.

H = 1/2 gt^2. (as u = 0.)

H = 1/2 ×10× 4

H=20 meters.

So,height of the building is 20 meters.

$\boxed{\boxed{H = 20 metres}}$

Case-3.

average speed for 2 seconds = ?.

Average speed = Total distance/Total time taken.

A.V = 20/2

A.V = 10 m/s.

So,average speed is 10m/s.

$\boxed{\boxed{Average speed =10m/s}}$