Answer:
Step-by-step explanation:
If we look at the general equation of position in time with constant accelerations
s = s₀ + v₀t + ½at²
and compare to our equation
h = 5 + 20t + -16t²
we can see that ½a = -16, therefore gravity is -32 ft/s²
that gravity will reduce the initial velocity to zero in
t = v₀/g = 20/32 = 0.625 s
at that time the ball will be
h = 5 + 20(0.625) + -16(0.625)² = 11.25 ft above the ground.
We could also find the time of maximum height by realizing the velocity at that moment would be zero.
As the slope of the position/time chart is velocity at any instant, taking the derivative of our equation and setting it equal to zero gives us
20 - 32t = 0
t = 0.625 s
same result.