Question

The height of an object t seconds after it is thrown from a height of h feet is modelled by the equation h(t) = -16t^2 + vt + h. If the ball is thrown from a point 6 feet above ground with an initial velocity, v, of 30 feet per second, how long will it take for the ball to hit the ground???​

Answer 1

Answer:

At 0.713 second the ball will hit the ground

Step-by-step explanation:

The height of an object t seconds after it is thrown from a height of h feet is modeled by the equation:

$h(t) = -16t^2 + vt + h$

The ball is thrown from a point 6 feet above ground with an initial velocity, v, of 30 feet per second, equation becomes:

$h(t) = -16t^2 + 3t + 6$

It is required to find the time when the ball hits the ground. At that point, h(t) =0.

So,

$-16t^2 + 3t + 6=0$

Above is a quadratic equation, the value of time is given by :

t = -0.526 s and t = 0.713 seconds

So, at 0.713 second the ball will hit the ground.