Question

A body is thrown up with a speed 49 m/s. It travels 5 m in the last second of its upward journey. If the
same body is thrown up with a velocity 98 m/s, how much distance (in m) will it travel in the last second
of its upward journey.
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Answer 1

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Answer 2

Answer:

The body is thrown up it travels 4.9 m in last second of upward journey.

Explanation:

Given that,

Speed = 49 m/s

Distance = 5 m

Speed of other body = 98 m/s

The velocity at t is zero and at last second is v =u.

So, the time is

$t = t-(t-1)$

$t = t-t+1=1\ sec$

We need to calculate the velocity

Using first equation of motion

$v = u+gt$

Put the value in the equation

$0=u-\times9.8\times1$

$u=9.8\ m/s$

We need to calculate the distance

Using third equation of motion

$v^2+u^2=2gs$

Put the value into the formula

$s=\dfrac{v^2+u^2}{2g}$

Put the value into the formula

$s=\dfrac{0+9.8^2}{2\times9.8}$

$s=4.9\ m$

Hence, The body is thrown up it travels 4.9 m in last second of upward journey.