Standing from the top of a platform. Frank
shot an arrow vertically into the air at an
initial velocity of 118 ft/s. The height of the
arrow, h, at t seconds is modeled by the equation h = -16t^2 + 118t + 15. How many seconds will take the arrow to reach the
ground?

Here, 118 ft/s is the initial velocity and 15 feet is the height of the platform. We need to find How many seconds will take the arrow to reach the ground. When it reaches the ground, h(t) = 0

[tex]-16t^2 + 118t + 15=0[/tex]

The above is a quadratic equation. The solution of the above equation is given by :

[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]

Here, b = 118, a = -16 and c = 15

So,

[tex]t=\dfrac{-118\pm \sqrt{(118)^2-4(-16)(15)} }{2\times (-16)}\\\\t=-0.125\ \text{and}\ 7.5\ s[/tex]

Time can't be negative.

So, it will take 7.5 seconds to reach the ground.

Standing from the top of a platform. Frank shot an arrow vertically into the air at an initial velocity of 118 ft/s. The height of the arrow, h, at t seconds is modeled by the equation h = -16t^2 + 118t + 15. How many seconds will take the arrow to reach the ground?

Respuesta :

Otras preguntas

Standing from the top of a platform. Frank
shot an arrow vertically into the air at an
initial velocity of 118 ft/s. The height of the
arrow, h, at t seconds is modeled by the equation h = -16t^2 + 118t + 15. How many seconds will take the arrow to reach the
ground?