## Saturday, July 20, 2013

### Differentiation - Rate Problem, 3

Category: Differential Calculus, Physics, Mechanics

"Published in Suisun City, California, USA"

If a ball is thrown vertically upward with a velocity of 80 ft/sec, then its height after t seconds is

(a) What is the maximum height reached by the ball?
(b) What is the velocity of the ball when it is 96 ft. above the ground on its way up? On its way down?

Solution:

If a ball is thrown in an upward position, then the height of a ball from the ground after t seconds is

To get the velocity of a ball in an upward position, take the derivative of the above equation with respect to time t, as follows

To maximize the height of a ball, equate the above equation to zero or set v = 0, we have

(a) Therefore, the height of a ball is

(b) If a ball is reached to 96 ft. from the ground in an upward position, then the velocity is calculated as follows

Substitute s = 96 ft. to the above equation, we have

Equate each factor to zero and the roots are t = 2 secs and t = 3 secs.

Since the time required to reach the maximum height is 2.5 secs, therefore we have to choose t = 2 secs.

Therefore, the velocity of a ball in an upward direction is

When a ball is reached to its maximum height, then the velocity of a ball in a downward direction at s = 96 ft. from the ground is

Where t = 3 secs (another root of a quadratic equation) after a ball is reached to its maximum height and the ball is about to fall when s = 96 ft. from the ground.

The velocity of a ball is negative because it is a downward direction after it reached to its maximum height. Also, the acceleration which is -16 = ½ g or g = - 32 ft/sec2 in the equation is negative because a ball is thrown upward. g is the acceleration due to gravity of an object in earth. g has different values depending in the planets and moon as well.