Question

a body is thrown up with a velocity 29.23 m/s distance travelled in last second of upward motion is

Answer 1

distance = speed/time n therefore the dist is 29.23m/s

Answer 2

Given that,

Velocity = 29 .23 m/s

We need to calculate the time

Using equation of motion

$v=u-gt$

Where, u = velocity

g = acceleration due to gravity

t = time

Put the value into the formula

$0=29.23-9.8\times t$

$t=\dfrac{29.23}{9.8}$

$t=2.98\ sec$

We need to calculate the distance travelled in last second

Using equation of motion

$S_{n}=ut-\dfrac{1}{2}g(2n-1)$

Where, g = acceleration due to gravity

t = time

Put the value into the formula

$S_{n}=29.23-\dfrac{9.8}{2}(2\times2.98-1)$

$S_{29.23}=4.9\ m$

Hence, The distance travelled in last second is 4.9 m

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Topic : distance travelled

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