Please help me solve this: If an object is projected upward with an initial velocity of 112 feet per second from a height h of 128 feet, then its height t seconds after it is projected is defined by the equation h=-16t^2 + 112t + 128. How many seconds after it is projected will it hit the ground?
To find the time at which the object hits the ground, we need to determine when the height of the object is equal to zero. In this case, the equation for the height of the object is given by:
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.
just solve for t when the height is zero:
-16t^2 + 112t + 128 = 0
just a normal quadratic equation.