The height of an object projected upward from ground level is given by h=-16t squared + 128t. When will the object be 240 feet above the ground? (Please provide steps/explanation to solve this problem.)
for this question, replace h = 240 into the given equation and solve for t.
240 = -16tsquared + 128t
-240 -240
0=-16tsquared+180t-240
solve for t by using quadratic formula or by regrouping.
goodluck
Thanks for the help. :)
To find when the object will be 240 feet above the ground, we need to solve the equation h = -16t^2 + 128t = 240.
Step 1: Rewrite the equation:
-16t^2 + 128t = 240
Step 2: Move all terms to one side of the equation:
-16t^2 + 128t - 240 = 0
Step 3: Divide the entire equation by -8 to simplify it:
2t^2 - 16t + 30 = 0
Step 4: Factor the equation:
(2t - 6)(t - 5) = 0
Step 5: Set each factor equal to zero and solve for t:
2t - 6 = 0 or t - 5 = 0
Solving the first equation:
2t = 6
t = 3
Solving the second equation:
t = 5
Therefore, the object will be 240 feet above the ground at two different times: t = 3 seconds and t = 5 seconds.
To find the time at which the object will be 240 feet above the ground, we can set the equation h = 240 feet and solve for t.
The given equation is h = -16t^2 + 128t. Substitute h = 240 into this equation:
240 = -16t^2 + 128t
Next, move all terms to one side to obtain a quadratic equation:
16t^2 - 128t + 240 = 0
We can simplify the equation by dividing through by 8:
2t^2 - 16t + 30 = 0
Now, let's solve this quadratic equation for t. There are multiple ways to do this, but one common method is factoring. However, this equation does not easily factor, so we will use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 2, b = -16, and c = 30. Plugging these values into the formula, we have:
t = (-(-16) ± √((-16)^2 - 4 * 2 * 30)) / (2 * 2)
Simplifying further:
t = (16 ± √(256 - 240)) / 4
t = (16 ± √16) / 4
Now, we have two possible values for t:
1. t = (16 + √16) / 4
t = (16 + 4) / 4
t = 20 / 4
t = 5
2. t = (16 - √16) / 4
t = (16 - 4) / 4
t = 12 / 4
t = 3
Therefore, the object will be 240 feet above the ground at two different times: t = 3 and t = 5.