Question

A ball is thrown from a height of 49 meters with an initial downward velocity of 2 m/s. The ball's height (in meters) after t seconds is given by the following.

h = 49 - 2t -
`5t^(2)`

How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

The height of the ball at ground level would be 0 feet.

Set h to 0 and solve for t:

Rewrite the equation as

-5t^2 -2t +49 = 0

Factor -1 out of the equation to get:

5t^2 + 2t -49 = 0

Use the quadratic formula:

-b +/- sqrt(b^2 -4(ac) / 2a

Where a = 5, b = 2 and c = -49

The quadratic formula becomes:

-2 +/-sqrt(2^2 – 4(5*-49) / (2*5)

Simplify to get :

T = -2 +/-sqrt246 / 10

T = 2.936 and – 3.336

Since the time has to be a positive value t = 2.936 seconds.

Rounded to the nearest hundredth = 2.94 seconds.