17. At what speed must an object be launched vertically upward for the object to reach a maximum height of 16 meters above its starting position.
(Do not include units, do not include a sign, use 3 sig figs) *

Explanation:

When starting at a height of 0, and initial velocity v, the equation of projectile motion is

y = -4.9x^2 + v*x

Where x is the time in seconds and y is the height at time x.

The vertex (h,k) of this parabola is the highest point. In general the vertex has its x coordinate at h = -b/(2a), so,

h = -b/(2a)

h = -v/(2*(-4.9))

h = v/(9.8)

Plug this into the first equation shown above. Also, plug in y = 16. Solve for v.

So we get the following.

y = -4.9x^2 + v*x

16 = -4.9(v/9.8)^2 + v*(v/9.8)

16 = (-5/98)v^2 + (5/49)v^2

16 = (-5/98)v^2 + (10/98)v^2

16 = (-5/98+10/98)v^2

16 = (5/98)v^2

(5/98)v^2 = 16

v^2 = 16*(98/5)

v^2 = 313.6

v = sqrt(313.6)

v = 17.708754896943

v = 17.7

The starting velocity must be roughly 17.7 meters per second.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.