Question 164030
Well, your answer is correct!
But you didn't get it from the equation you posted because there is an error in the equation.
The general form of the equation for this kind of problem is:
{{{h(t) = -16t^2+v[0]t+h[0]}}} where: h = height, in feet, t = time, in seconds, {{{v[0]}}} = initial upward velocity of the object, and {{{h[0]}}} = the initial height of the object.
For the situation in your problem, the equation should be:
{{{h(t) = -16t^2+150t+5}}}
To find the time at which Charlie O'Brien's baseball returns to the ground, you would set h(t) = 0 and solve for t, so...
{{{-16t^2+150t+5 = 0}}} Solve this quadratic equation by using the quadratic formula:{{{t = (-b+-sqrt(b^2-4ac))/2a}}}
In this problem, a = -16, b = 150, and c = 5, so making the substitutions, we get:
{{{t = (-150+-sqrt(150^2-4(-16)(5)))/2(-16)}}} Simplifying this, we get:
{{{t = (-150+-sqrt(22500-(-320)))/-32}}}
{{{t = (-150+-sqrt(22820))/-32}}}
{{{t = (-150+151)/-32}}} or {{{t = (-150-151)/-32}}}
{{{t = -0.3125}}} or {{{t = 9.4}}} Discard the negative value as the time should be positive.
It would take 9.4 seconds for Charlie O'Brien's baseball to return to the ground.