With s(t) measured in feet and t in seconds, the equation s(t)= -16t^2 + 32t expresses the height of a ball above the ground after t seconds if it were thrown upward with a velocity of 32 ft/sec.
(a) What is the maximum height of the ball?
(b) After how many seconds will the maximum height occur?
(b) recall that the vertex of a parabola is at t = -b/2a. In this case, that's 32/32 = 1.
(a) just plug in t=1.
WRONG
To find the maximum height of the ball, we need to find the vertex of the quadratic equation s(t) = -16t^2 + 32t.
(a) The vertex of a quadratic equation in the form of y = ax^2 + bx + c is given by the formula x = -b/2a. In this case, a = -16 and b = 32. Plugging these values into the formula, we have:
t = -b/2a
t = -32/(-2*16)
t = 1
To find the maximum height, substitute t = 1 into the equation:
s(1) = -16(1)^2 + 32(1)
s(1) = -16 + 32
s(1) = 16 feet
Therefore, the maximum height of the ball is 16 feet.
(b) The maximum height occurs at t = 1 second.