A waterfall has a height of 1000 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 24.

Question

A waterfall has a height of 1000

feet. A pebble is thrown upward from the top of the falls with an initial velocity of 24

feet per second. The? height, h, of the pebble after t seconds is given by the equation

h= 16t^2+24t+1000.
How long after the pebble is thrown will it hit the? ground?

Answer 1

First, let us correct the equation.  It should say

 

h(t) = -16t² + 24t + 1000  

 

(The minus sign was missing.  The minus sign indicates the acceleration of gravity is in the downward direction, which is compatible with the given initial velocity being positive in the upward direction.)

 

The pebble hits the ground when  h = 0

 

h = -16t² + 24t + 1000 = 0

 

Divide by -4 to simplify this a little and get

 

4t² - 6t - 250 = 0

 

Use the quadratic equation to find t.

 

t = [1/(2*4)] {6 ±√[(-6)² - 4(4)(-250)]}

 

= (1/8) (6 ± 63.53)

 

Use the positive square root to get a positive time.

 

t = (1/8) (6 + 63.53) s = 8.691 s

 

The pebble hits the ground 8.691 s after it is tossed.  You can test the result by plugging this time back into the original h(t) and see that h(t) = 0.

Answer 2

Mason,

 

Actually, the height function is incorrect. It should be

 

h(t) = −16t² + 24t + 1000.

 

The negative sign is due to the influence of gravity pulling the pebble downward as it travels through the air.

 

To answer your question set this function equal to zero and solve for t.

 

−16t² + 24t + 1000 = 0

 

2t² − 3t − 125 = 0  (divide both sides by the common factor of −8)

 

Solve using the quadratic formula:

 

t = (3 ± √[(-3)² − 4(2)(-125)])/(2·2) = (3 ± √1009)/4.

 

If you compute these two roots, you get the values t ≈ 8.69 and t ≈ −7.19. The second one obviously doesn't make sense, so the correct answer is 8.69 seconds, rounded to two decimal places.