A ball is thrown upward from the ground with a velocity of 32 ft/sec. Determine the velocity, max height and time.

Question

A ball is thrown upward from the ground with a velocity of 32 ft/sec. Determine

a) the average velocity of the ball on [3/4, 5/4] seconds;

b) the instantaneous velocity at t=3/4sec and at t=5/4 sec;

c) the maximum height attained by the ball; and

d) the time it takes the ball to return to the ground.

Answer 1

h(t) = -16t^2 + 32t

h'(t) = -32t + 32 = 0

t= 32/32 = 1 second to reach maximum height,

and t=2 seconds to reach the ground

max height = h(1) =-16(1)^2 +32 = 16 feet

velocity at t=3/4 = h'(3/4) = -32(3/4) +32 = 8 feet per second, going up

velocity at t =5/4 is h'(5/4) = -32(5/4) +32 = -8 feet per second. It's going down

the height function is a downward opening parabola with vertex = (1, 16)

axis of symmetry is x=1

which explains why the tiime to reach the ground is exactly 2 seconds

and the 8 feet per second velocities have the same absolute value for t=3/4 as t =5/4

(0, 0) and (2, 0) are the x intercepts, equidistant from the axis of symmetry

average velocity on the interval t=3/4 to t=5/4 is zero.

It's going up half the time, from 3/4 to 1, then down from 1 to 5/4 at the same speed, just opposite direction, which average out to zero