Question

If a particle is thrown with velocity more than
10 m/s vertically upward, then the distance
travelled by the particle in last second of its
ascent is​

Answer 1

Answer:

Please refer to the above pic attached.

Hope it helps you !

Answer 2

Answer:

s=4.9m.

Explanation:

given,

u=o

t=1sec

v=10m/s

a=g=9.8m/s^2.

s=?

$s=ut*1/2gt^2$

s=(0*10)*1*2*9.8*(1)^2

s=4.9*1

s=4.9m.

HOPE IT'S HELPS YOU... ✌