Question

A

1) Maya launches a baseball directly upward at 64 feet per
second from a platform 8 feet high. Write the function for
the height of this object at any given time (t seconds).
h =
When will the object reach it's maximum height?
What will that maximum height be?

Answer 1

• 2 seconds
• 72 feet

Explanation:

The usual equation used for the vertical component of ballistic motion is ...

h(t) = -16t² +v₀t +h₀

where v₀ is the initial upward velocity, and h₀ is the initial height. Units of distance are feet, and units of time are seconds.

Your problem statement gives ...

v₀ = 64 ft/s

h₀ = 8 ft

so the equation of height is ...

h(t) = -16t² +64t +8

__

For quadratic ax² +bx +c, the axis of symmetry is x=-b/(2a). Then the axis of symmetry of the height equation is ...

t = -64/(2(-16)) = 2

The object will reach its maximum height after 2 seconds.

The height at that time will be ...

h(2) = -16(2²) +64(2) +8 = 72

The maximum height will be 72 feet.