A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 127 feet per second from an initial height of 114 feet off the ground, then the height of the projectile, h, in feet, t seconds after its shot is given by the equation:
h=-16t^2+127t+114
Find the two points in time when the object is 151 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).
Answer: The object is 151 feet off the ground at the following times:
To find the two points in time when the object is 151 feet above the ground, we need to solve the equation:
-16t^2 + 127t + 114 = 151
Let's rearrange the equation to standard quadratic form:
-16t^2 + 127t + 114 - 151 = 0
-16t^2 + 127t - 37 = 0
Now we can solve this quadratic equation. We can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 127, and c = -37.
t = (-127 ± √(127^2 - 4*(-16)*(-37))) / (2*(-16))
Simplifying this expression, we get:
t = (-127 ± √(16129 - 2368)) / -32
t = (-127 ± √13761) / -32
Now let's calculate the two possible values for t:
t ≈ (-127 + √13761) / -32 ≈ 4.7077 seconds
t ≈ (-127 - √13761) / -32 ≈ 0.7323 seconds
Therefore, the object is 151 feet off the ground at approximately 0.73 seconds and 4.71 seconds.