Acannon launches a cannonball from a height of 3 feet with an upward velocity of 40 feet per second. the function h=-16t^2+40t+3 gives the height "h" (in feet) of the cannonball after "t" seconds.

a. after how many seconds is the cannonball 10 feet above the ground?

b. what is the maximum height of the cannonball?

answer with the correct unit!

A)  We want to know when h=10 so

10=-16t^2+40t+3

16t^2-40t+7=0

Using the quadratic formula:

t=(40±√1152)/32 

t≈0.189 and 2.311 seconds

So it is at 10 ft twice, about 2/10 of a second when it is rising and at 2 3/10 of a second when it is falling...

b.

The maximum height of the ball is when velocity is equal to zero, or dh/dt=0

dh/dt=-32t+40, dh/dt=0 when 32t=40, t=40/32=5/4=1.25 seconds

h(1.25)=28ft

So the maximum height of the ball occurs after 1.25 seconds and reaches a height of 28 feet.

The answer of this question is 672 .