Problema Solution
A ball is thrown from the top of a building. Its height, in feet, t seconds later is given by h(t)=-16t^2=32t+50.
1.)What is the maximum height of the ball?
2.)How long before the ball gits the ground?
Answer provided by our tutors
1.)What is the maximum height of the ball?
we need to find the maximum of the parabolic function h(t)=-16t^2 + 32t+50
h max = c - (b^2)/(4a), where a = -16, b = 32, c = 50
h max = 50 - (32^2)/(4*(-16))
h max = 66 ft
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the maximum height of the ball is 66 ft
2.)How long before the ball gits the ground?
when the ball hits the ground -16t^2 + 32t+50 = 0 and t>0
by solving the quadratic equation we find
t = 3.03 s
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the ball will hit the ground after 3.03 seconds.