SOLUTION: A softball player throws a ball into the air at an initial velocity of 44 feet per second. The player’s hand is 5 feet from the ground at the time the ball is released. Use th

Question 1189509: A softball player throws a ball into the air at an initial velocity of 44 feet per second. The player’s hand is 5 feet from the ground at the time the ball is released.
Use the formula −16t2 + vt + c = 0.
t = time in seconds the ball is in the air
v = initial velocity in feet per second
c = the initial distance from the ground
h = the distance above the ground in feet
Q. How many seconds will the ball stay in the air? Round your answer to the nearest hundredth second.
a. 1.09
b. 1.59
c. 1.82
d. 2.86
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A softball player throws a ball into the air at an initial velocity of 44 feet per second. The player’s hand is 5 feet from the ground at the time the ball is released.
Use the formula −16t2 + vt + c = 0.
t = time in seconds the ball is in the air
v = initial velocity in feet per second
c = the initial distance from the ground
h = the distance above the ground in feet
Q. How many seconds will the ball stay in the air? Round your answer to the nearest hundredth second.
a. 1.09
b. 1.59
c. 1.82
d. 2.86
=====================================
h(t) = −16t^2 + vt + c = 0
-16t^2 + 44t + 5 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=2256 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.10929276087974, 2.85929276087974. Here's your graph:

t = ~2.86 seconds

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