Question

a ball is tossed into the air at 30 feet per second from a height of 8 feet. how long will it take the ball to reach the ground? use the formula ( s = -16t^2 + v_0 t + s_0 ), where ( v_0 ) = initial velocity, ( s_0 ) = initial altitude, and ( s ) = altitude in feet after ( t ) seconds. the approximate time it took was (square) seconds. (round to the nearest tenth.)

Explanation:

Step1: Identify known values

We know that when the ball reaches the ground, the altitude \( s = 0 \). The initial velocity \( v_0 = 30 \) feet per second and the initial altitude \( s_0 = 8 \) feet. The formula is \( s=- 16t^{2}+v_0t + s_0 \). Substituting the known values, we get the quadratic equation:
\( 0=-16t^{2}+30t + 8 \)
We can rewrite this equation as \( 16t^{2}-30t - 8=0 \) (multiplying both sides by - 1 to make the coefficient of \( t^{2} \) positive).

Step2: Use quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). In our equation \( 16t^{2}-30t - 8 = 0 \), we have \( a = 16 \), \( b=-30 \), and \( c = - 8 \).
First, calculate the discriminant \( D=b^{2}-4ac \):
\( D=(-30)^{2}-4\times16\times(-8)=900 + 512=1412 \)
Then, find the square root of the discriminant: \( \sqrt{1412}\approx37.58 \)
Now, substitute into the quadratic formula:
\( t=\frac{-(-30)\pm37.58}{2\times16}=\frac{30\pm37.58}{32} \)
We have two solutions:
\( t_1=\frac{30 + 37.58}{32}=\frac{67.58}{32}\approx2.1 \)
\( t_2=\frac{30-37.58}{32}=\frac{-7.58}{32}\approx - 0.2 \)
Since time cannot be negative, we discard \( t_2 \).

Answer:

\( 2.1 \)