Question

1. A person drops the ball from the roof of a house. It takes 3 seconds for the ball to hit the ground. Calculate the velocity of the ball right before it hits the grounds.

2. A car at rest accelerates at 2 m/s^2. After 10 seconds what is the velocity of the car and how far has the car traveled?

3. Tasha throws a ball straight up in the air. After 1 second it reaches its maximum height then it free falls for 2 seconds. Calculate the maximum height of the ball and the velocity of the ball before it hits the ground.

Answer 1

Answer 1 :

Time taken by the ball to hit the ground (t) = 3 s .

Velocity before hitting the ground (v) = v .

Initial velocity of the ball (u) = 0 m/s .

We know that acceleration due to gravity (g) = 10 m/s² .

By Laws of Motion we know that :

v = u + gt

⇒ v = 0 m/s + 10 m/s² × 3 s

⇒ v = 0 m/s + 30 m/s

⇒ v = 30 m/s

The velocity of the ball before hitting the ground is 30 m/s .

Answer 2 :

Acceleration of the car (a) = 2 m/s² .

Time taken by the car (t) = 10 s .

Initial velocity of the car (u) = 0 m/s

Let the final velocity be v .

Let the distance travelled be S.

By Laws of Motion we know that :

v = u + at

⇒ v = 0 m/s + 2 m/s² × 10 s

⇒ v = 0 m/s + 20 m/s

⇒ v = 20 m/s

By Laws of Motion we know that :

v² = u² + 2 a S

⇒ ( 20 m/s )² = u² + 2 a S

⇒ 400 m²/s² = ( 0 m/s )² + 2 × 2 m/s² × S

⇒ 400 m²/s² = 4 m/s² × S

⇒ S = 400 m²/m × 1/4

⇒ S = 100 m

The distance travelled is 100 m .

The velocity is 20 m/s .

Answer 3 :

Time taken to free fall  (t) = 2 s

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 10 m/s² .

By Laws of motion :

S = ut + 1/2 gt²

⇒ S = 0 + 1/2 × 10 m/s² × ( 2 s )²

⇒ S = 1/2 × 10 m × 4

⇒ S = 20 m

By Laws of motion we know that :

v = u + gt

⇒ v = 0 m/s + 10 m/s² × 2 s

⇒ v = 0 m/s + 20 m/s

⇒ v = 20 m/s

The velocity is 20 m/s .

The maximum height of the ball is 20 m .

Answer 2

ANSWER

[1]Given,

-Time taken by the ball to hit the ground is 3 seconds. (s)

-The initial velocity of the ball will be 0 m/s.(u)

(since the ball was at rest before)

-Let the velocity of the ball when it hits the ground be v m/s

-Also, acceleration produced due to gravity is 10 m/$s^2$(g force).

This will be acceleration here.

From the laws of kinematics, we have >

° v = u+at

=> v = u+at

⇒ v = 0 + 10 × 3

⇒ v = 0 m/s + 30 m/s

⇒ v = 30 m/s

Hence,

the velocity of the ball before hitting the ground will be 30-meter per second.

{ 2 }Given,

-Acceleration of the car is 2 m/$s^2$.(a)

-Time is taken by the car for acceleration is 10 s. (t)

-The initial velocity of the car will be

0 m/s (u)

(as the car was initially at rest)

-Let the final velocity of the car be v.

and the distance traveled be s.

Now, we know that the first Equation of kinematics is=>

° v = u + at

⇒ v =0 + 2 × 10

⇒ v = 0 m/s + 20 m/s

⇒ v = 20 m/s

So,

velocity of the car is 20 m/s

By using the second equation of kinematics we have=>

$v^2$-$u^2$= 2 a×s

=>$20^2$ - $0^2$= 2 ×2×s

=> 400 = 4 × S

=> 400 = 4 × S

=>S = $\dfrac{400}{4}$

=> S = 100 m

Hence,

the car traveled a distance of100 m with velocity 20 m/s.

[3]Given,

-Time of free fall of ball= 2 s(t)

-The initial velocity of ball= 0m/s(u)

-The acceleration produced due to gravity =10 m/$s^2$(g)

This will be acceleration.

Now, by using the second law of kinematics we have=>

S = ut + $\dfrac{1}{2}$a$t^2$

=>S = ut + $\dfrac{1}{2}$ g$t^2$2

=> S = 0 + $\dfrac{1}{2}$×10×$2^2$

=> S =$\dfrac{1}{2}$ × 40

=> S =20 m.

So

the distance traveled by car is 20m

Now, again by using laws of kinematics we have =>

°v = u + at

=> v = u+gt

=> v = 0 + 10 × 2

=> v = 0 + 20

=> v = 20 m/s

Hence the final velocity of the ball is 20m/s.

Remember

3 equations of kinematics are =>

1)v = u + at

2)s = ut + $\dfrac{1}{2}$ ×at^2

3)$v^2$-$u^2$= 2 a×s