Question

A ball thrown vertically upwards returns to the thrower after 4 seconds find
i) the velocity with which it was thrown up
ii) the maximum height it reaches
iii) its position after 3 seconds.
​

Answer 1

Answer:

Refer the picture given below

Please mark me as the brainliest

Answer 2

Given :

• Final velocity, v = 0 m/s
• Time taken, t = 2s
• Acceleration due to gravity, g = -10 m/s

To find :

• (i)the velocity with which it was thrown up
• (ii) the maximum height it reaches
• (iii) its position after 3 seconds.

Solution :

(i)The velocity with which it was thrown up :-

Using the formula,

★v = u + gt

Putting values,

=> 0 = u+(−10)×2

=> 0 - u = - 20

=> u=20m/s

Therefore,

The velocity with which it was thrown up is 20 m/s .

__________________

ii)The maximum height it reaches :-

Using the formula,

★h = ut + 1/2 gt^2

Putting values,

=>h=20×2 + 1/2 × -10 ×2×2

=>h=40−20

=>h=20m

Therefore, 

The maximum height it reaches is 20 m.

___________________

iii)Its position after 3 seconds:-

Time taken for the downward journey :

=> Time taken to reach maximum height - time at which it falls down.

=> 4 seconds - 3 seconds

=> 1 seconds.

Using the formula,

★h = ut + 1/2 gt^2

Putting values,

=>h = 0 + 1/2 ×10×1×1

=>h = 10/2

=>h = 5m

Position after 3s :

=> 20 - 5

=> 15 m

Therefore, 

Its position after 3 seconds is 15 m.

_____________________