A projectile launched straight up has height s (in feet) at time t (in seconds) where s(t) = -16t^2 + 140t + 75. What is the initial height and the maximum height of the projectile?
Can you please answer this? please? thank you!

Question

anonymous · Answer

The initial height is the height at t=0.  That shouldn't be too hard to find.

The maximum height takes a little more figuring.  Notice that the function s(t) is a parabola that opens down.  That means the maximum height happens at the vertex of the parabola.  We find the t value of that point by letting t equal$$t=\frac{-b}{2a}=\frac{-140}{-32}=\frac{35}{8}$$From there, find the maximum height by plugging this value into the function.$$s(\frac{35}{8})=-16(\frac{35}{8})^2+140(\frac{35}{8})+75$$You should be able to calculate this.

anonymous · Answer

omg thank you so so much. I was so confused but you cleared it up so nicely. You're better than my math teacher! Thank you!

anonymous · Answer

No sweat.  Do math every day.

anonymous · Answer

alrighty! thank you so much!!