Answer:
Correct answer: A.) V = - 16.6 m/s down
Explanation:
Given:
V₀ = 3 m/s initial velocity
t = 2 seconds
g = 9.8 m/s²
V(t) = V(2) = ?
The movement described is a vertical upward shot
For velocity at any time is valid the next formula
V = V₀ - g · t
V = 3 - (9.8 · 2) = 3 - 19.6 = - 16.6 m/s down
Under condition that it has a enough drop height with respect to the ejection point.
God is with you!!!
Take the tiny bit of carbon dioxide and the tiny bit of water vapor out of the air,
and the rest of what you're breathing right now is a mixture of elements.
Answer:
E. 3h
Explanation:
We know that
u = 0 m/s.
velocity after t = 1s
v = u+gt = 0+9.81 x 1s= 9.81 m/s
distance covered in 1st sec
= =>> ut+0.5 x g x t²
=>>0 + 0.5x 9.81 x 1 = 4.90m
Let 4.90 be h
distance travelled in 2nd second will now be used
So velocity after t = 1s
=>>1 x t+ 0.5 x g x t²
=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90
So since h= 4.90
Then the ans is 3x h = 3h