An object is thrown upward from the top of a 160-foot building with an initial velocity of 144 feet per second. The height, h, of the object after t seconds is given by the equation h=-16t^2+144t+160. When will the object hit the ground?
The object will hit the ground when the time is _____ seconds.
To find the time when the object hits the ground, we need to determine when the height, h, of the object is equal to 0.
The equation given is: h = -16t^2 + 144t + 160.
So we set h = 0, and solve the equation for t:
0 = -16t^2 + 144t + 160.
To solve this quadratic equation, we can use either factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = -16, b = 144, and c = 160. Substituting these values into the quadratic formula, we have:
t = (-(144) ± √((144)^2 - 4(-16)(160))) / (2(-16)).
Simplifying further:
t = (-144 ± √(20736 + 10240)) / (-32).
t = (-144 ± √(30976)) / (-32).
t = (-144 ± √(176)) / (-32).
Taking the positive square root:
t = (-144 + √(176)) / (-32).
Calculating further:
t ≈ (-144 + 13.26) / (-32).
t ≈ (-130.74) / (-32).
t ≈ 4.08 seconds (rounded to two decimal places).
Therefore, the object will hit the ground after approximately 4.08 seconds.
To find when the object will hit the ground, we need to determine the time when the height (h) of the object is equal to 0. In this case, the equation for the height of the object is h = -16t^2 + 144t + 160, where h represents the height in feet and t represents the time in seconds.
So, we need to solve the equation -16t^2 + 144t + 160 = 0 for t.
To do this, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a, b, and c are the coefficients of the equation (-16t^2 + 144t + 160 = 0).
In this case, a = -16, b = 144, and c = 160.
Substituting these values into the quadratic formula, we get:
t = (-(144) ± √((144)^2 - 4(-16)(160))) / (2(-16)).
Simplifying further:
t = (-144 ± √(20736 + 10240)) / (-32),
t = (-144 ± √(30976)) / (-32),
t = (-144 ± 176) / (-32).
Now, we can calculate the two possible solutions:
t1 = (-144 + 176) / (-32),
t1 = 32 / (-32),
t1 = -1.
t2 = (-144 - 176) / (-32),
t2 = -320 / (-32),
t2 = 10.
Therefore, the object will hit the ground after 10 seconds.
So, the answer is: The object will hit the ground when the time is 10 seconds.
when it hits the ground, h = 0
so solve:
-16t^2 + 144t + 160 = 0
divide each term by -16
t^2 - 9t - 10 = 0
easy to solve, use factoring.