Question
Please give the answer in exact number! No approx!
An object is shot vertically upward from the ground with an initial velocity of 288 ft/sec. At what rate is the velocity decreasing? Give units. Explain why the graph of velocity of the object against time (with upward positive) is a line. The graph of velocity is a line because the acceleration is decreasing at a constant rate. The graph of velocity is a line because the velocity is constant. The graph of velocity is a line because the acceleration is increasing at a constant rate. The graph of velocity is a line because the velocity is decreasing at a constant rate. The graph of velocity is a line because the velocity is increasing at a constant rate. Using the starting velocity and your answer to part (b), find the time at which the object reaches the highest point. Use your answer to part (c) to decide when the object hits the ground. Graph the velocity against time. Mark on the graph when the object reaches its highest point and when it lands. Find the maximum height reached by the object by considering an area on a graph. Now express the velocity as a function of time. Find the greatest height by antidifferentiation.
Explanation / Answer
a) 32 feet/second^2
b) because velocity is decreasing at a constant rate
c) at max height velocity = 0
so 288 = 32t (as it is decreasing at a constant rate )
t= 9 sec
d) 9 sec as the velocity increases at the same rate
e) its A ... at t=18 sec it reaches the ground
f)area = 2 * (0.5*288*9) = 1296 ft
height= 1296 ft
g) from graph slope is -a y intercept is 288
so v = -a*t +288
a= 32 ft/s^2
integrating from 0 to 9
-0.5aT^2 +288t
-0.5*32*9*9 + 288*9
= 1296 ft