Question

a ball is thrown vertically up with a velocity of 50m/s. find the distance on last second is

Answer 1

u=50m/s
v=0
g= 9.8(-10m/s) approx.
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v²=u²+2as

s=v²-u²/2a

s= 0²-50²/2×(-10)

s= -50²/ -20

s= -(2500)/ -20

s=2500/20

s=125m
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Answer 2

Given ---
Initial velocity (u) = 50 m/s
Final velocity (v) = 0 m/s....(as it will stop after covering some distance)
Acceleration due to gravity (g) = -9.8 m/s² (negative because the ball is going against the gravitational force)

To find ---
Distance (s)

Solution ---
We know, by third equation of motion, that

$2as = {v}^{2} - {u}^{2}$
$= > 2gs = {v}^{2} - {u}^{2}$
$= > s = \frac{ {v}^{2} - {u}^{2} }{2g}$
$= > s = \frac{ {0}^{2} - {50}^{2} }{2 \times ( - 9.8)}$
$= > s = \frac{ - 2500}{ - 19.6}$
$= > s = \frac{2500}{19.6}$
$= > s = 127.5$
Hence, the ball will travel 127.5 m before accelerating back.

Hope it'll help.. :-)