SOLUTION: the height in feet for a ball thrown upward at 48 feet persecond is given by s(t)=-16t2+48t, where t is the time in seconds after the ball is tossed. what is the maximum height t
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Question 10369: the height in feet for a ball thrown upward at 48 feet persecond is given by s(t)=-16t2+48t, where t is the time in seconds after the ball is tossed. what is the maximum height that the ball will reach?
You can put this solution on YOUR website! This is the equation for a parabola that opens downward. The quadratic equation is already in standard form: where x = t, a = -16, b = 48, and c = 0.
The maximum height of the thrown object will be at the vertex (the maximumum value) (h, t) of the parabola.
The t-coordinate (equivalent to the x-coordnate) of the vertex (this is the time at which the object reaches its maximum height) is given by where a = -16 and b = 48.
so, at t= 3/2 secs, the object is at its maximum height.
To find the value of the maximum height at t = 3/2 secs, substitute t = 3/2 into the original quadratic equation and solve for h.
As a check, you could take the first derivative of the quadratic equation and set it to zero to find the value of t at the maximum height. = secs, same as previous answer for the time the object reaches maximum height.
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