Question

a body is dropped from a height of 320 M the acceleration due to gravity is 10 M per square how a how long does it take to reach the ground be what is the velocity with which it will strike the ground

Answer 1

time of decent =root under 2h÷g=
$\sqrt{2 \times 320 \div 10}$
8sec
u =0 ,a=10m/s^2,t=8 sec
a
hence a =(v-u)÷t =(v-0)÷8 =10 hence v=80m/s

Answer 2

S=320m
a=10m/sec^2
u=0m/sec
t=?
v=?
we know that
S=it+1\2at^2
320=(0)t+1/2(10)t^2
320*2=(10)t^2
640/10=t^2
√64=t
t=4sec
now v= ?
we know that
v=u+at
v=0(10)4
v=40 m/sec
answer