A ball is thrown into the air with an initial upward velocity of 48ft/s its height h in feet after t seconds is given by the function h = -16t^2 + 48t + 64

6 seconds
4 seconds
7 seconds
5 seconds

Sorry, I forgot to put part of the question.

A ball is thrown into the air with an initial upward velocity of 48ft/s its height h in feet after t seconds is given by the function h=-16t^2 + 48t + 64 After how many seconds will the ball hit the ground?

when it hits the ground, the height is zero, right?

so just solve
-16t^2 + 48t + 64 = 0
or, dividing by -16,
t^2 - 3t - 4 = 0
sure looks like t=4

To find the height of the ball at different times, you can substitute the given time values into the equation h = -16t^2 + 48t + 64 and solve for h.

Let's find the height of the ball after 6 seconds:

h = -16(6)^2 + 48(6) + 64
h = -16(36) + 288 + 64
h = -576 + 288 + 64
h = -224 + 64
h = -160

Therefore, the height of the ball after 6 seconds is -160 feet.

Now, let's find the height of the ball after 4 seconds:

h = -16(4)^2 + 48(4) + 64
h = -16(16) + 192 + 64
h = -256 + 192 + 64
h = -64 + 64
h = 0

Therefore, the height of the ball after 4 seconds is 0 feet.

Next, let's find the height of the ball after 7 seconds:

h = -16(7)^2 + 48(7) + 64
h = -16(49) + 336 + 64
h = -784 + 336 + 64
h = -448 + 400
h = -48

Therefore, the height of the ball after 7 seconds is -48 feet.

Finally, let's find the height of the ball after 5 seconds:

h = -16(5)^2 + 48(5) + 64
h = -16(25) + 240 + 64
h = -400 + 240 + 64
h = -160 + 64
h = -96

Therefore, the height of the ball after 5 seconds is -96 feet.

To summarize:
- After 6 seconds: Height = -160 feet
- After 4 seconds: Height = 0 feet
- After 7 seconds: Height = -48 feet
- After 5 seconds: Height = -96 feet

all or any of the above, depending on the question ...