Question

(If there is more than one answer, use the "or" button.)Round your answer(s) to the nearest hundredth.A ball is thrown from a height of 141 feet with an initial downward velocity of 21 ft/s. The ball's height h (in feet) after t seconds is given by the following.h = 141 - 21t - 16t ^ 2How long after the ball is thrown does it hit the ground?

Answer 1

Given:

`h=141-21t-16t^2`

To get the time the ball hit the ground, it hits the ground when the height is zero.

Hence,

`\begin{gathered} At\text{ h = 0;} \\ h=141-21t-16t^2 \\ 0=141-21t-16t^2 \\ 141-21t-16t^2=0 \\ 16t^2+21t-141=0 \end{gathered}`

To solve for t, we use the quadratic formula.

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{where;} \\ a=16,b=21,c=-141 \\ t=\frac{-21\pm\sqrt[]{21^2-(4*16*-141)}}{2*16} \\ t=\frac{-21\pm\sqrt[]{441+9024}}{32} \\ t=\frac{-21\pm\sqrt[]{9465}}{32} \\ t=(-21\pm97.288)/(32) \\ t_1=(-21+97.288)/(32)=(76.288)/(32)=2.384\approx2.38 \\ t_2=(-21-97.288)/(32)=(-118.288)/(32)=-3.6965\approx-3.70 \end{gathered}`

Since time can't be a negative value, we pick the positive value of t.

Therefore, to the nearest hundredth, it takes 2.38 seconds for the ball to hit the ground.