Question

A ball is thrown from an initial height of 4 feet with an initial upward velocity of 33 ft/s. The ball's height h (in feet) after t seconds is given by the following.
h= 4 + 33t - 16t^2
CAN SOMEONE PLEASE PLEASE HELP ME

Find all values of t for which the ball's height is 20 feet.


Round your answer(s) to the nearest hundredth.

Answer 1

Answer:

Two methods both tell us the ball is at 20 feet height at 0.783 seconds going up and 1.28 seconds on the way down.

Step-by-step explanation:

The question can be answered by either calculation or by graphing.

Calculation:

Set the equation to h = 20 feet and solve with the quadratic equation.

h= 4 + 33t - 16t^2

4 + 33t - 16t^2 = 20

- 16t^2 + 33t + 4 = 20

- 16t^2 + 33t - 16 = 0

I find solutions of 0.779 and 1.28 seconds.  

Graphing:

See the attached graph.

The graph shows solutions of 0.783 and 1.28 seconds.

Both methods show the ball to be at 20 feet height at 0.783 seconds going up and 1.28 seconds on the way down.