Question

A ball is thrown vertically upwards with a velocity of 20m/s. Calculate the maximum height reached by ball. How long does it remain in the air?

Answer 1

Explanation:

Given:

• Initial velocity (u) = 20 m/s
• Final velocity (v) = 0 m/s

To find:

• Maximum height or distance reached by ball (s) = ?
• How long it will remain in the air (t) = ?

Solution:

☆ We know, acceleration (a) = v - u/t

(putting values)...

--> 10 = 0 - 20/t

--> 10 = 20/t

--> 10 × 20 = t

--> 200 s = t

Time = 200 seconds.

☆ We know Newton's second equation of motion i.e, s = ut + 1/2×at^2

(putting values)...

--> s = 20 × 200 + 1/2 × 10 × (200)^2

--> s = 4,000 + 5 × 40,000

--> s = 4,000 + 2,00,000

--> s = 2,04,000 m

Maximum height = 2,04,000 meters.

Hope it helped you dear...

Answer 2

$\red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

It is given that :

• Initial velocity of ball = 20m/s
• Final velocity = 0 at maximum height
• Distance or height reached = ?
• Time in air = ?

$\blue{\mid{\underline{\overline{\textbf{Coming to question :-}}}\mid}}$

In this question a ball is projected upwards with a velocity of 20m/s , the maximum height it reaches is not known and the time it spent in air we have to calculate it.

Now, if we talk about acceleration of ball then it is :

$\bigstar \: a = \frac{v - u}{t}$

$\mapsto \: a = \frac{0 - 20}{t}$

In this case the acceleration acting on ball vertically is the acceleration due to gravity = 9.8

$\mapsto \: 9.8 = \frac{ - 20}{t}$

$\mapsto \: t = \frac{ - 20}{9.8} = 2.0$

Therefore time spent in air = 2s

Now according to third equation of motion :

${v}^{2} = {u}^{2} + 2as$

$\bigstar \: s = \frac{ {v}^{2} - {u}^{2} }{2a}$

After inputting the known values we get :

$\mapsto \: s = \frac{0 - 400}{9.8 \times 2}$

$s = \frac{400}{19.6}$

$\bigstar \: s = 20$

Therefore the total distance traveled by ball is 20m.

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BY SCIVIBHANSHU

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