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Question

Suppose that a ball is dropped from the upper observation deck of a building, 300 m above the ground.

(a) What is the velocity of the ball after 3 seconds?
(b) How fast is the ball traveling when it hits the ground?

SOLUTION We will need to find the velocity both when t = 3 and when the ball hits the ground, so it's efficient to start by finding the velocity at a general time t = a. Using the equation of motion s = f(t) = 4.9t² , we have

 

= lim (h->0)     (f(a + h) - f(a)) / h 

= lim (h->0)      (4.9( (a+h)²   - 4.9a²)/h 


= lim (h->0)       (4.9(a² + 2ah + h² - a²))/h


= lim (h->0)        (4.9(2ah + h2 - a2))/h


= lim (h->0)         (4.9(?????????????)/h 

=                          ????????????

 

The velocity after 3 s is v(3) = 9.8( ??????? ) = ??????????????

Answer 1

Hi Mia,

 

Velocity is the change in position with respect to time which is the first derivative of the position function. 

 

For the position function   s = f(t) = 4.9t² ,  the first derivative is  f'(t) = 9.8t   Therefore after 3 sec.,

 

                                         f'(3) = 4.9(3)  = 29.4 m/sec.  

 

To find the velocity when it hits the ground, we must first determine how long it takes to hit the ground. We nedd to set the function  

                                        f(t) = 4.9t² = 300 and solve for t.  In this case t = √300/4.9 ≈ 76.7 m/sec.