Physics, asked by ashwinilakhan6, 2 months ago

A ball of mass 40 g is thrown upward. It reaches to the maximum height of 80 m. At what height its kinetic energy reduced by 40%?​

Answers

Answered by nirman95
15

Given:

A ball of mass 40 g is thrown upward. It reaches to the maximum height of 80 m.

To find:

Height at which KE reduced by 40%.

Calculation:

First of all, let initial velocity be u :

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

  \implies {0}^{2}  =  {u}^{2}  - 2(10)(80)

  \implies  {u}^{2}   = 1600

  \implies  u   = \sqrt{1600}

  \implies  u   = 40 \: m {s}^{ - 1}

Now, initial KE :

KE =  \dfrac{1}{2} m {u}^{2}

 \implies KE =  \dfrac{1}{2}  \times 0.04 \times  {(40)}^{2}

 \implies KE = 32 \: joule

Now, KE is reduced by 40%.

  • So, final KE = 60% (Initial KE)

 \implies KE_{f}  =60\% \times  32

 \implies KE_{f}   = 19.2

 \implies \frac{1}{2}  \times 0.04 \times  {v_{f}}^{2}   = 19.2

 \implies v_{f}  = 30.98 \: m {s}^{ - 1}

Now, required height will be :

 { v_{f}}^{2}  =  {u}^{2}  - 2gh

  \implies 960 = 1600  - 2(10)h

  \implies \: 20h = 640

  \implies \: h = 32 \: m

At height 32 metres, KE will be reduced by 40%.

Answered by mad210215
1

Given:

mass of ball m = 40g

maximum height Hmax = 80 m

To find:

height at which kinetic energy is reduced by 40% h =?

Explanation:

  • A ball of mass 40 g is thrown upward & reaches the maximum height of 80 m.

         i.e. m = 40 g

         Hmax = displacement  = S = 80 m

  • Hence at maximum height, the velocity of a ball becomes zero.

        i.e. final velocity v = 0 m/s

  • First, we have to find the initial velocity with which a ball is thrown upward.  
  • By the kinematical law of equation

v^2 = u^2 + 2as

0 = u^2 + 2 (-9.81) (80)

  • After solving above equation,

u = 39.6181 m/s

  • The kinetic energy of ball is calculated as:

K.E = 0.5 × mu^2

     = 0.5 × 0.040 × 39.6181 (39.6181)

K.E = 31.3918 J

  • Now, 40% of K.E = 0.40 × 31.3918 = 12.5567 J

12.5567 = 0.5 × mv^2

⇒ v = 25.05 m/s

  • from the law of the kinematical equation, we can find the height

v^2 = u^2 + 2as

(25.05)(25.05) = (39.6181)(39.6181) + 2(-9.81) s

s = 48.01 m

Hence at a height of 48.01 m, the kinetic energy reduced by 40% of a ball.

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