Question

The cricket baĺl has an initial velocity of 20 m/s vertically upwards what height does the ball reach before it stops to fall back to the ground

Answer 1

Answer

☞Given -

$\sf u = 20 m/s$

$\sf a = g = - 10 m/s²$

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$a is acceleration due to gravity.

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☞To find -

Height reached by ball before falling back to earth - Maximum height reached by the ball $\longrightarrow$ s

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☞Formula used -

3rd equation of motion $\longrightarrow$ v² = u² + 2as

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☞Solution -

As the ball reached the maximum height and then stopped to come back , the final velocity of an object is 0

$\sf u = 20 m/s$

$\sf a = -10 m/s²$

$\sf v = 0 m/s$

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Substituting the value in third equation of motion -

$\sf v² = u² + 2as$

$\longrightarrow$$\sf 0 = 20 \times 20 - 2 \times 10 \times s$

$\longrightarrow$$\sf 20s = 400$

$\longrightarrow$$\sf s = 20m$

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☞Additional information -

v = u + at

s = ut + 1/2 at²

Snth = u + a/2 ( 2n - 1)

where

$\longrightarrow$v is final velocity.

$\longrightarrow$u is initial velocity.

$\longrightarrow$s is distance travelled.

$\longrightarrow$a is acceleration.

$\longrightarrow$t is time taken.

$\longrightarrow$n is second

Thanks