a ball is thrown into the air at a speed of 96 feet/sec. The equation that expresses the height of the ball as a function of time is: h(t) = -16t2 + 96t + 5
How long does the ball stay in the air after it is thrown? (Round answer to the hundreths of a second)
To find out how long the ball stays in the air after it is thrown, we need to find the time at which the height function, h(t), equals zero.
The height function is given by the equation: h(t) = -16t^2 + 96t + 5.
To find the time at which the ball hits the ground, we set h(t) equal to zero:
0 = -16t^2 + 96t + 5
Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.
In this case, let's use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
In our equation, a = -16, b = 96, and c = 5.
Plugging the values into the quadratic formula, we get:
t = (-96 ± √(96^2 - 4(-16)(5))) / (2*(-16)).
Simplifying further, we have:
t = (-96 ± √(9216 + 320)) / -32
t = (-96 ± √9536) / -32.
To find the two possible solutions, we calculate both the positive and negative values:
t1 = (-96 + √9536) / -32
t2 = (-96 - √9536) / -32.
Calculating further, we get:
t1 ≈ -0.17
t2 ≈ 5.92.
Since time cannot be negative in this context, we discard t1 as a solution. Therefore, the ball stays in the air for approximately 5.92 seconds after it is thrown.
Rounding the answer to the hundredths of a second, the ball stays in the air for approximately 5.92 seconds.