Answer:
b. v = 0, a = 9.8 m/s² down.
Explanation:
Hi there!
The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?
Let´s take a look at the height function:
h(t) = h0 + v0 · t + 1/2 g · t²
Where
h0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.
Another way to see it (without calculus):
When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.
If this case could ever happen, the speed would follow from this formula:
with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."
The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)
5.6•10^5 so it’s to the power of positive 5
