While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone s
ubsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
impact speed:
18.1151
m/s
How much time is the stone in the air?
elapsed time:
You final velocity is correct but not to the correct number of significant digits. The actual answer should be 18.1 m/s. We will use that to find the total time the stone was in the air in the equation:
v = v₀ + at
18.1 = 6.63 + (-9.81)t and
11.5 = -9.81t so
t = 1.17 seconds.
We know this was how long the stone was in the air (as compared to the time that the stone took to reach its max height or some other height) because we used the velocity with which the stone hit the ground to find the total time the rock was in the air before it hit the ground going at that velocity.
You can solve this using Kepler's 3rd law. T is proportional to r^1.5. T(J) = T(E)*5.2^1.5 = 11.86 years. <span>You can also solve it the hard way: Find M (sun's mass) and r(Jupiter's orbital radius) in m. Then use Kepler's equation T = 2pi*r^1.5/sqrt(GM).</span>
