Problema Solution
A pistol is accidentally discharged vertically upward at a height of 3 feet above the ground. If the bullet has an initial muzzle velocity of 200 feet per second, what maximum height will it reach before it starts to fall to the ground?
Answer provided by our tutors
s(t) = -16t^2 + v0t + s0
s(t) = the height after t seconds
v0 = 200 ft/s is the initial velocity
s0 = 3 ft is the initial height
t = is the time
s(t) = -16t^2 + 200t + 3
We need to find the maximum of the above parabolic equation:
s max = c - b^2/(4a)
In our case a = -16, b = 200, c = 3
s max = 3 - 200^2/(4*(-16))
s max = 628 ft
The bullet will reach maximum height of 628 ft.