Question

A body is thrown up with a velocity 29,23 m/sec distance traveled in last second of upward
motion is in metre is​

Answer 1

Explanation:

$\huge\underline\mathfrak{Answer}$

GIVEN:-

Initial velocity,u = 29.23 m/s

TO FIND:-

Distance travelled in the last second of its upward motion

PROCEDURE

We know that ,the distance covered by a vertically projected body in the last second of its upward motion = distance it covers in the first second of its downward motion

We know that,

Distance covers in nth second = g(n - 1/2) {for freely falling body}

So,

Distance travelled in last second = g(1-1/2)

= g(1/2)

= g/2

we know that g=9.8 m/s^2

= 9.8/2

= 4.9m

HOPE IT HELPS YOU

Answer 2

Answer:

Explanation: always start from top point bcz speed is zero there so all equations bcome easy .