Answer: [tex]2.5\ seconds[/tex]
Step-by-step explanation:
The complete exercise is: "Jennifer hit a golf ball from the ground and its movement follows the equation [tex]h(t)=-16t^2+100[/tex], where t is the time in seconds, and h(t) is the height of the ball at a given time "t". Find the time that it took the golf ball to hit the ground."
The quadratic function is:
[tex]h(t)=-16t^2+100[/tex]
Since the ground is at [tex]h(t)=0[/tex], you need to substitute this into the function:
[tex]0=-16t^2+100[/tex]
Now you must solve for "t". The steps are:
- You need to subtract 100 from both sides:
[tex]0-100=-16t^2+100-100\\\\-100=-16t^2[/tex]
- Now you must divide both sides of the equation by -16:
[tex]\frac{-100}{-16}=\frac{-16t^2}{-16}\\\\6.25=t^2[/tex]
- Applying the square root both sides of the equation, you get:
[tex]\±\sqrt{6.25} =\sqrt{t^2}\\\\t_1=2.5\\\\t_2=-2.5[/tex]
