In physics, it is shown that the height s of a ball thrown straight up with an

Question:

In physics, it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft sec from a rooftop 112 feet high is

s = s(t) = -16 t2 + 96t + 112

where t is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground. 

(a) When does the ball strike the ground? That is, how long is the ball in the air? 

(b) At what time t will the ball pass the rooftop on its way down? 

(c) What is the average speed of the ball from t = 0 to t = 2?

(d) What is the instantaneous speed of the ball at time t? 

(e) What is the instantaneous speed of the ball at t = 2?

(f) When is the instantaneous speed of the ball equal to zero? 

(g) What is the instantaneous speed of the ball as it passes the rooftop on the way down? 

(h) What is the instantaneous speed of the ball when it strikes the ground?

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question

Precalculus

ISBN: 978-0321716835

9th edition

Authors: Michael Sullivan

Question Posted: