just solve for t when the height is zero:
-16t^2 + 112t + 128 = 0
just a normal quadratic equation.
-16t^2 + 112t + 128 = 0
just a normal quadratic equation.
h = -16t^2 + 112t + 128
Setting h = 0, we have:
-16t^2 + 112t + 128 = 0
To solve this quadratic equation, we can either factor, complete the square, or use the quadratic formula. In this case, factoring might not be a straightforward option, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 112, and c = 128. Substituting these values into the quadratic formula, we get:
t = (-112 ± √(112^2 - 4(-16)(128))) / (2(-16))
Simplifying further:
t = (-112 ± √(12544 + 8192)) / (-32)
t = (-112 ± √(20736)) / (-32)
We can simplify the square root:
t = (-112 ± 144) / (-32)
Now, we have two possible values for t: one with the positive square root and one with the negative square root:
t1 = (-112 + 144) / -32
t2 = (-112 - 144) / -32
Calculating each value:
t1 = 32 / -32
t1 = -1
t2 = -256 / -32
t2 = 8
We ignore the negative value since time cannot be negative in this context. Therefore, the object will hit the ground after 8 seconds.