A girl stands on the edge of her dorm roof and throws a ball into the air. The ball's height above the ground after t seconds is given by h equals negative 16 t squared plus 64 t plus 80h=−16t2+64t+80. When will the ball hit the ground?
find t when h=0.
If you divide through by -16, then just solve
t^2 - 4t - 5 = 0
To find out when the ball will hit the ground, we need to determine the time at which the height of the ball, denoted by "h," equals zero.
The given equation gives the height of the ball as a function of time:
h = -16t^2 + 64t + 80
When the ball hits the ground, its height is zero. So we can set the height equation equal to zero and solve for t:
-16t^2 + 64t + 80 = 0
To solve this quadratic equation, we can either factor it, complete the square, or use the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, the coefficients are:
a = -16, b = 64, and c = 80
Substituting these values into the quadratic formula, we get:
t = (-64 ± √(64^2 - 4(-16)(80))) / (2*(-16))
Simplifying the equation, we have:
t = (-64 ± √(4096 + 5120)) / (-32)
t = (-64 ± √9216) / (-32)
t = (-64 ± 96) / (-32)
Now, let's consider both cases separately:
1. t = (-64 + 96) / (-32)
t = 32 / (-32)
t = -1
2. t = (-64 - 96) / (-32)
t = -160 / (-32)
t = 5
The ball cannot have a negative time value, so the only valid solution is t = 5.
Therefore, the ball will hit the ground after 5 seconds.