43.9k views
2 votes
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.

1 Answer

1 vote

Answer:

246 ft is the maximum height

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

User Apchester
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories