The height h, in feet, of a projectile t seconds after launch is modeled by the equation h=32t-16t^2.

Complete question is;

The height h, in feet, of a projectile t seconds after launch is modeled by the equation h = 32t – 16t². How long after launch does the projectile return to the ground?

2 seconds

Step-by-step explanation:

We are told that the height function is given by;

h = 32t – 16t²

Now, the velocity will be the derivative of the height equation.

Thus;

v(t) = dh/dt = 32 - 32t

Now, for projectiles, at max height, v = 0.

Thus;

32 - 32t = 0

32t = 32

t = 32/32

t = 1 s

It will take the same time it took to get to Max height as it will to get to ground from that height.

Thus, total height = 1 + 1 = 2 seconds

answer: $201

step-by-step explanation: to get the answer, you first get 3 nieces that he sold and multiply it by 67 and get $201, so he earned $201 selling his nieces.

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length