College Now Math

Question

An object is thrown upward at a speed of 126 feet per second by a machine from a height of 16 feet off the ground. The height of the object after t seconds can be found using the equation s(t)= -16t^{2 +} v0^t +s0, where v0 is the initial velocity and s0 is the initial height. Give all numerical answers to 2 decimal places.

1. when will the object reach its maximum height?
2. What is its maximum height?
3. when will the object reach the ground?

Answer 1

maybe there's a misprint. this problem would be so much easier if initial velocity was 128, not 126

s(t) =-16t^2 + 128t + 16 is the height equation then

take the derivative and set =0

-32t + 128 = 0

t = 4 seconds for maximum height

plug that value into the first equation s(4) = 272 feet

for hitting the ground set s(t) =0, use the quadratic formula

t=4+sqr17 = about 8.12 seconds, about twice the time to reach the max height

but if the initial velocity was 126, then max height is reached in 3 15/16 seconds = 3.9375 seconds.

Plug that into the original equation and solve for s

It just makes the calculations messier