Question

A cannon ball is launched into the air with an upward velocity of 91 feet

Answer 1

Given the model of the projectile as

`h(t)=-16t^2+91t+3`

Step-by-step explanation

Number 1: At height 21, we will have

`\begin{gathered} -16t^2+91t+3=21 \\ -16t^2+91t-18=0 \\ solving\text{ with quadratic formula} \\ t_(1,\:2)=(-91\pm √(91^2-4\left(-16\right)\left(-18\right)))/(2\left(-16\right)) \\ t_(1,\:2)=(-91\pm √(7129))/(2\left(-16\right)) \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ t=(91-√(7129))/(32),\:t=(91+√(7129))/(32) \\ t=0.21,\:t=5.5 \end{gathered}`

Answer

t=0.21 seconds

Number 2:

To find the time the projectile will take to touch the ground, we will equate the function to zero.

`\begin{gathered} -16t^2+91t+3=0 \\ solving\text{ with quadratic formula} \\ t_(1,\:2)=(-91\pm √(91^2-4\left(-16\right)\cdot \:3))/(2\left(-16\right)) \\ t_(1,\:2)=(-91\pm √(8473))/(2\left(-16\right)) \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ t=-(-91+√(8473))/(32),\:t=(91+√(8473))/(32) \\ t=-0.03277,\:t=5.720 \end{gathered}`

Anwetr t=5.7 seconds