if an object is thrown vertically upward with an initial velocity of v from an original position of s, the height h at any time t is given by: h=-16t^2+vt+s (where h and s are in ft t is in seconds and v is in ft/sec) If a rock is thrown upward from a height of 100ft, with an initial velocity of 32ft/sec, solve for the time that it takes for it to hit the ground (when h=0)

just plug in the numbers and solve for t in

-16t^2 + 32t + 100 = 0
t = 3.69

To solve for the time it takes for the rock to hit the ground, we need to find the value of t when h = 0.

Given that h = -16t^2 + vt + s, we substitute h = 0 and the given values for v and s:

0 = -16t^2 + 32t + 100

Now, we have a quadratic equation. To solve for t, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -16, b = 32, and c = 100. Substituting these values:

t = (-32 ± √(32^2 - 4(-16)(100))) / (2(-16))

Simplifying the equation further:

t = (-32 ± √(1024 + 6400)) / (-32)

t = (-32 ± √7424) / (-32)

t = (-32 ± √(64 * 116)) / (-32)

t = (-32 ± 8√29) / (-32)

Now we have two possible solutions for t:

t1 = (-32 + 8√29) / (-32)

t2 = (-32 - 8√29) / (-32)

Simplifying further:

t1 ≈ -0.84 seconds

t2 ≈ 6.84 seconds

Since time cannot be negative in this situation, the rock will hit the ground after approximately 6.84 seconds.