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)
a ball is thrown upward with initial velocity of 96 ft/sec from the top of a 100 ft. bridge. Determine the time that it takes for the ball to get to a height of 200 ft.
Round your answers to 2 decimals, separated by a comma.
Plug the given values into the given Eq and solve for t.
To determine the time it takes for the ball to reach a height of 200 ft, we can substitute the given values into the equation for height:
h = -16t^2 + vt + s
Let's plug in the values:
h = 200 ft
v = 96 ft/sec (initial velocity)
s = 100 ft (original position)
200 = -16t^2 + 96t + 100
Now we have a quadratic equation. To solve for t, we can rearrange the equation by moving all terms to one side:
0 = -16t^2 + 96t + 100 - 200
Simplifying:
0 = -16t^2 + 96t - 100
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 96, and c = -100. Substituting these values into the quadratic formula:
t = (-96 ± √(96^2 - 4(-16)(-100))) / (2(-16))
Calculating the square root:
t = (-96 ± √(9216 -(-6400))) / (-32)
t = (-96 ± √(9216 + 6400)) / (-32)
t = (-96 ± √15616) / (-32)
The square root of 15616 is approximately 124.97, so:
t = (-96 ± 124.97) / (-32)
Now we can solve for t by considering both the positive and negative solutions:
t1 = (-96 + 124.97) / (-32)
t2 = (-96 - 124.97) / (-32)
Calculating:
t1 = 28.97 / (-32) ≈ -0.90 seconds
t2 = -220.97 / (-32) ≈ 6.91 seconds
Since time cannot be negative in this context, we can conclude that it takes approximately 6.91 seconds for the ball to reach a height of 200 ft.