Question

a ball is thrown from the top row of seats in a stadium. the function $h(t) = -16t^2 + 128t + 50$ gives the height, $h$, in feet, of the ball $t$ seconds after it is thrown. how long will it be before the ball hits the ground? round your answer to the nearest whole number. the ball will hit the ground after \boxed{} seconds.

Explanation:

Step1: Set \( h(t) = 0 \)

We need to find when the ball hits the ground, so set \( h(t)=0 \). The function is \( h(t)= - 16t^{2}+128t + 50 \), so we have the quadratic equation:
\( -16t^{2}+128t + 50=0 \)
Multiply both sides by - 1 to simplify:
\( 16t^{2}-128t - 50 = 0 \)
We can use the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for a quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a = 16 \), \( b=- 128 \), \( c=-50 \).

Step2: Calculate the discriminant

First, calculate the discriminant \( D=b^{2}-4ac \):
\( D=(-128)^{2}-4\times16\times(-50) \)
\( D = 16384+3200 \)
\( D=19584 \)

Step3: Apply the quadratic formula

Now, find \( t \) using the quadratic formula:
\( t=\frac{-(-128)\pm\sqrt{19584}}{2\times16}=\frac{128\pm\sqrt{19584}}{32} \)
Calculate \( \sqrt{19584}\approx139.94 \)
We have two solutions:
\( t_1=\frac{128 + 139.94}{32}=\frac{267.94}{32}\approx8.37 \)
\( t_2=\frac{128-139.94}{32}=\frac{-11.94}{32}\approx - 0.37 \)
Since time cannot be negative, we discard \( t_2 \). Rounding \( t_1 \) to the nearest whole number gives \( t\approx8 \).

Answer:

8