Two balls of equal masses are thrown upwards along the same vertical direction at an interval of 2 second, with the same initial velocity of 39.2m/s. Calculate the height where the collide.
Answers
Answered by
0
u1=3.92m/s=u2
t=2s
m1=m2
v=0
a=(v-u)/t
=(0-3.92)/2
=-3.92/2
=-1.96
h=ut+1/2 at^2
=3.92×2+1/2×-1.96×2^2
=7.84×1/2×-1.96×4
=7.84×1/2×7.84
=30.7328
t=2s
m1=m2
v=0
a=(v-u)/t
=(0-3.92)/2
=-3.92/2
=-1.96
h=ut+1/2 at^2
=3.92×2+1/2×-1.96×2^2
=7.84×1/2×-1.96×4
=7.84×1/2×7.84
=30.7328
Answered by
5
Heya!!!
Here is your answer,
Hope it helps you,
Thank you.
Here is your answer,
Hope it helps you,
Thank you.
Attachments:
Similar questions