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 dropped from the top of a 200 ft building, solve for the time that it takes for it to hit the ground(when h=0) Hint: Initial velocity when dropped is 0. Round your answer to 2 decimals. Enter: seconds

Question

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 dropped from the top of a 200 ft building, solve for the time that it takes for it to hit the ground(when h=0) Hint: Initial velocity when dropped is 0. Round your answer to 2 decimals. Enter: seconds

Answer 1

since v=0,

h = -16t^2 + 200

0 = -16t^2 + 200
t = 3.54

Answer 2

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

Given:
h = -16t^2 + vt + s
Initial velocity when dropped, v = 0
Height of the building, s = -200 ft (negative because it is below the reference point)

Substituting the given values into the equation, we have:
h = -16t^2 + 0t - 200

Now, we can set h equal to 0 and solve for t:
0 = -16t^2 - 200

To solve this quadratic equation, let's first divide both sides by -16 to simplify the equation:
t^2 + 12.5 = 0

Next, move 12.5 to the other side of the equation by subtracting it from both sides:
t^2 = -12.5

To solve for t, we need to take the square root of both sides:
t = √(-12.5)

However, the square root of a negative number is not a real number. This means that the rock will never hit the ground based on the given equation.

Please note that this result may be due to limitations in the equation or the given values. Make sure you have provided accurate information and check if there are any missing details.