Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
D. Light as light is a wave
Cl2 is the answer. Hope this helps you.
The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.
Considering also the formula for Gravitational Potential Energy GPE = mgh
For the bird on the ground, h =0, therefore GPE = m*9.8*0 = 0
For that on the tree = mgh = m*9.8*h
Of course the one on the tree has a value greater than zero.