Respuesta :
Given:
The height of the projectile shooting up into the air is given as,
[tex]h\text{ = -16t}^2+128t[/tex]Required:
The time t required for the projectile to return to its starting position.
Explanation:
When the projectile starts from its initial position and comes back to the same position then the height becomes zero.
Equating the given equation to zero.
[tex]\begin{gathered} -16t^2+\text{ 128t = 0} \\ \end{gathered}[/tex]Calculating the roots of the given quadratic equation.
[tex]\begin{gathered} 16t^2-128t\text{ = 0} \\ t(16t-128)\text{ = 0} \end{gathered}[/tex]On simplifying further,
[tex]\begin{gathered} t\text{ = 0 or 16t - 128 = 0} \\ t\text{ = 0 or t = }\frac{128}{16} \\ t\text{ = 0 or t = 8} \end{gathered}[/tex]Therefore,
At t = 0 seconds, the projectile was shot into the air. After 8 seconds the projectile returned back to its original position.
Answer:
Thus the time taken by projectile to return back to its starting position is 8 seconds.
