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.
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Answer 1

Solution :-

Given :-

Ball returns to the ground after 4 seconds.

∴ The time taken by the ball to reach maximum height is 2s.

i)

Final velocity, v = 0 m/s

Time taken, t = 2s

Acceleration due to gravity, g = -10 m/s

$\boxed{\bf v = u+gt}$

$: \implies \sf 0 = u+ -10 \times 2$

$: \implies \sf \bf u = 20 m/s$

ii)

$\boxed{\bf h= ut+\dfrac{1}{2}gt^2}$

$: \implies \sf h = 20 \times 2+ \dfrac{1}{2} \times -10 \times 2 \times 2$

$: \implies \sf h = 40-20$

$: \implies \bf h = 20m$

iii)

The ball takes 2s to reach the maximum height.

After 3s, it starts to fall down.

Time taken for the downward journey = 4-3 = 1s

$: \implies \sf h = 0+ \dfrac{1}{2} \times 10 \times 1^2$

$: \implies \sf h = 5m$

Position after 3s = 20-5

                           = 15m

 

Answer 2

Answer:

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

Solution :

i) Given :

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

As we know that,

$\huge \bf \underline \blue{v = u + gt}$

Then,

$\implies \sf 0 = u + -10 \times 2$

$\implies \sf 0 - u = - 20$

$\implies \sf \bf u = 20 m/s$

Therefore, the velocity with which it was thrown up is 20 m/s .

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ii) Given :

• Final velocity, v = 0 m/s
• Time taken, t = 2s
• Acceleration due to gravity, g = -10 m/s
• Initial velocity = 20 m/s

As we know that,

${\purple{\boxed{\large{\bold{{h= ut+\dfrac{1}{2}gt^2}}}}}}$

Then,

$\implies \sf h = 20 \times 2+ \dfrac{1}{2} \times - 10 \times 2 \times 2$

$\implies \sf h = 40 - 20$

$\implies \sf h = 20m$

Therefore, the maximum height it reaches is 20 m .

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iii) The ball takes 2s to reach the maximum height.

Then, after 3s, it starts to fall down.

Time taken for the downward journey,

=> 4 seconds - 3 seconds

=> 1 seconds.

Given :

• Final velocity, v = 0 m/s
• Time taken, t = 1s
• Acceleration due to gravity, g = 10 m/s

Then,

$\implies \sf h = 0+ \dfrac{1}{\cancel{2}} \times {\cancel{10}} \times 1^2$

$\implies \sf h = 5m$

Then,

$\leadsto$ Position after 3s,

=> 20 - 5

=> 15 m

Therefore, its position after 3 seconds is 15 m.

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