Math Question (Velocity)

Question

A ball is thrown vertically upward from the ground with an initial velocity of 114 ft per sec. Use h = −16t2 + v0t + h0. Determine what the maximum height will be. (Round your answer to two decimal places.)

Answer 1

h(t) = -16t^2 + vot + ho

vo = initiial velocity, ho = initial height, -16 = a/2 or g/2, the effect of gravity

take the derivative and set equal to zero, then solve for time t

dh/dt = h'(t) = -32t + 114 = 0

t = 114/32 = 3.5625 seconds to reach max height

plug that t value into the original equation to solve for maximum height

max height = h(3.5625) = -16(3.5625)^2 + 114(3.5625) + ho

= -203.0625 + 406.125 + ho

= 203.0625 + ho

= 203.06 feet rounded to 2 decimal places with ho = 0 as it starts at ground level

actually 32.17405 feet per second per second is a closer value for the acceleration due to gravity, at sea level

it gets rounded off to 32, and with the general formula h(t) = -16t^2 +vot + ho where 16 is half 32

graphically the height equation is a downward opening parabola with vertex=maximum = about (3.56, 203.06)