Question 1014451
The baseball is hit with an initial vertical velocity of 80 feet per second when it is 3 feet off the ground.
:
a. Write an equation that gives the height (in feet) of baseball as a function of the Tim (in seconds) since it was hit.
let t = time of the flight of the baseball
then
f(t) = height of the ball after t time; -16t^2 is downward force of gravity
f(t) =-16t^2 + 80t + 3
:
b. After how many seconds does the ball reach a height of 99 feet?
-16t^2 + 80t + 3 = 99
-16t^2 + 80t + 3 - 99 = 0
-16t^2 + 80t - 96 = 0
simplify, divide equation by -16
t^2 - 5t + 6 = 0
Solve for t, factors to
(t-3)(t-2)
two solutions
t = 2 seconds reaches 99 ft on the way up
and
t = 3 seconds descends to 99ft on the way down
looks like this, green line is 99ft
{{{ graph( 300, 200, -4, 8, -10, 120, -16x^2+80x+3, 99) }}}
:
c. Does the ball reach 99 feet more than once? Justify your answer.
Initial velocity formula: -16t^2+vt+s
We answered this
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