Question

a teacher uses a strong slingshot to release an object from the top of a school. the function $a(t) = -16t^2 + 128t + 50$ gives the approximate altitude, in feet, of the object $t$ seconds after it is released. how long will it be before the object hits the ground? round to the nearest second.

Explanation:

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

We need to find when the object hits the ground, so set the altitude function \( a(t)= -16t^{2}+128t + 50\) equal to 0: \( -16t^{2}+128t + 50=0\). Multiply both sides by -1 to simplify: \(16t^{2}-128t - 50 = 0\).

Step2: Use quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 16\), \(b=- 128\), \(c=-50\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-128)^{2}-4\times16\times(-50)=16384 + 3200=19584\). Then, \(t=\frac{128\pm\sqrt{19584}}{32}\). \(\sqrt{19584}\approx139.94\). So we have two solutions: \(t_1=\frac{128 + 139.94}{32}\approx\frac{267.94}{32}\approx8.37\) and \(t_2=\frac{128-139.94}{32}\approx\frac{- 11.94}{32}\approx - 0.37\). Since time cannot be negative, we take the positive solution. Rounding \(8.37\) to the nearest second gives 8.

Answer:

8 seconds