Answer:
t = 4 seconds
Step-by-step explanation:
The height of the projectile after it is launched is given by the function :
[tex]h(t)=-16t^2+32t+128[/tex]
t is time in seconds
We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0
So,
[tex]-16t^2+32t+128=0[/tex]
The above is a quadratic equation. It can be solved by the formula as follows :
[tex]t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
Here, a = -16, b = 32 and c = 128
[tex]t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s[/tex]
Neglecting negative value, the projectile will land after 4 seconds.
