Question

40. A person throws a ball vertically up returns to him after 6 sec. Find.
(a) the velocity with which it was thrown up.
the agree with the
(b) the maximum height it reaches.
(c) its position after 4 s.​

Answer 1

★Solution:-

The time taken by the ball to reach max height ( time of ascent ) will be equal to time taken by the ball to return to it's original position ( time of descent )

Let ,

Time of ascent = time of descent = t

Total time taken by the ball = 6 s ( given )

➝T = t + t

➝ T = 2t

➝ t = T /2

➝ t = 6 /2

➝ t = 3 sec

• The time taken by the ball to reach maximum height = 3 sec

• The final velocity of the ball at the highest point = 0 m/s
• Acceleration due to gravity = - 10 m/s²( as the ball is moving against gravity )

(a) Initial velocity of the ball

As the ball moves with uniform acceleration throughout it's motion we can use the first kinematic equation in order to find u

➽ v = u + gt

here ,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• t = time

➽ 0 = u - 10 x 3

➽ u = 30 m /s

The initial velocity of the ball is 30 m/s

(b) Maximum Height

As the ball moves with uniform acceleration throughout it's motion we can use the second kinematic equation in order to find h

➽ h= ut +gt²/2

here,

• h = maximum height
• u = initial velocity
• g = acceleration due to gravity
• t = time

➽ h = 30 x 3 - 10 x 9/2

➽ h = 90 - 5 x4

➽ h = 90 - 45

➽ h = 45 m

The maximum height reached by the ball is 45 m

(c) Position after 4 sec

We know that after 3 sec the ball starts to move downwards

So first we need to find the distance covered by the ball in 1 second

• time = 1 sec

• Acceleration  due to gravity = 10 m/s²( as the ball is moving along gravity )

• Initial velocity = 0 m/s

As the ball moves with uniform acceleration throughout it's motion we can use the second kinematic equation in order to find h'

➽ h'= ut +gt²/2

➽ h'= ut +gt²/2

➽ h' = 0 x 1 + 10 x 1 / 2

➽ h' = 5m

The position of the ball after 4 sec = 45 - 5 = 40 m

Answer 2

Answer:

Given:-

→Initial velocity of the body=50m/s

→Acceleration due to gravity= -10m/s²

To find:-

→Maximum height reached by the body

→Time taken to reach maximum height

Solution:-

In this problem:-

•Acceleration due to gravity is given

-10m/s²,as the body is going against gravity

•Final velocity of the body at the highest point is zero,thus while solving this problem,we will take v=0

By using the 3rd equation of motion,we get:-

=>v²-u²=2gh

=>0-(50)²= 2(-10)h

=>-2500 = -20h

=>h = -2500/-20

=>h = 125m

Now,by using 1st equation of motion,we get:-

=>v=u+at

=>0=50+(-10)t

=>-50 = -10t

=>t = -50/-10

=>t = 5s

Thus:-

→Maximum height reached by the

body is 125m.

→Time taken to reach maximum

height is 5s.