Velocity is the rate of change in the height of the projectile. So, to find the velocity from the height function H(t) = 58t - 0.83t2, you need to find the derivative H'(t).
(a) Using the power rule, H'(t) = 58 - 1.66t. So at time t = 1, the velocity is 58 - 1.66(1) = 56.34 m/s.
(b) If t = a, we simply replace the t in the velocity function with a. So the velocity at t = a is 58 - 1.66a.
(c) When the arrow hits the moon, its height is 0. There is more than one way to do this. Method 1: Set H(t) equal to 0 and solve for t.
58t - 0.83t2 = 0
58t = 0.83t2
58 = 0.83t
t = 69.9 s
Method 2: Since this function has no constant term, at time 0 the height is also 0. In other words, it is launched from the ground. This being the case, this makes the parabola perfectly symmetrical. So if we know how long it takes to reach its peak, then it will take the same amount of time to return to the ground.
We can find when it reaches its peak by finding when the velocity is equal to 0.
58 - 1.66t = 0
58 = 1.66t
t = 34.9
Double that value, and that it is the time when it returns to the ground, at t = 69.9.
(d) The velocity when it hits the ground, or the impact velocity, can be found by plugging the time from part (c) into the velocity function. In other words, what is the velocity at time t = 69.9?
58 - 69.9*1.66 = -58 m/s.
You might think that you had a mistake in getting a negative number. But remember that velocity is both a speed and a direction, and the negative sign means that it is going the opposite direction from when it was launched.
This answer looks curiously familiar, doesn't it? It is the same velocity as the initial velocity, but in the opposite direction. This also makes sense in light of the perfect symmetry of the parabolic arc that the graph makes. You can also think of it in a common-sense way by reasoning that, whatever speed was lost as the arrow was rising to its peak altitude, it is all gained back as it returns and accelerates toward its original altitude.
