A ball is thrown straight up from the top of a building 124 ft tall with an initial velocity of 64 ft per second. The height s(t) (in feet) of the ball from the ground, at time t (in seconds), is given by s(t) = 124 + 64t − 16t2. Find the maximum height attained by the ball.

246 ft is the maximum height

Step-by-step explanation:

The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t² + bt + c, the vertex is located at t = - b / 2a. Hence for h given above the vertex in the question s(t) = 124 + 64t − 16t², is at t

t = -64/2(-16) = 64/32 = 2 seconds

Thus, 2 seconds after the object was thrown, it reaches its highest point (maximum value of h) which is given by

h = -16(2)² + 64 (2) + 124 = 246eet

Distributive Property (of multiplication)

Step-by-step explanation:

You distribute the equation of 2a(3a+1) into 6a^2+2a by multiplying