A baseball player hits a foul ball straight up in the air from a height of 4 feet off the ground with an initial velocity of 85 feet per second.
When is the ball 110 feet off the ground?
To find out when the ball is 110 feet off the ground, we need to determine the time it takes for the ball to reach that height.
We can use the kinematic equation:
h = h0 + v0t - (1/2)gt^2
Where:
- h is the height of the ball at time t
- h0 is the initial height of the ball (4 feet)
- v0 is the initial velocity of the ball (85 feet per second)
- g is the acceleration due to gravity (-32.2 feet per second squared)
- t is the time it takes for the ball to reach a certain height
In this case, we want to find the time when the ball is at a height of 110 feet, so we can rearrange the equation to solve for t:
110 = 4 + 85t - (1/2)(-32.2)t^2
Let's solve this equation step by step to find the time t.
To find out when the ball is 110 feet off the ground, we need to determine the time it takes for the ball to reach that height.
The motion of the ball can be described using the equation for vertical displacement:
y = y0 + v0*t - (1/2)gt^2
where:
y is the vertical displacement (height) of the ball
y0 is the initial height (4 feet off the ground)
v0 is the initial velocity (85 feet per second)
t is the time
g is the acceleration due to gravity (32 feet per second squared)
We can rearrange this equation to solve for time:
110 = 4 + 85*t - (1/2)*32*t^2
Simplifying the equation, we have:
0 = 16t^2 - 85t + 106
The equation is a quadratic equation, which can be solved using various methods such as factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 16, b = -85, and c = 106. Substituting these values into the quadratic formula, we have:
t = (85 ± √((-85)^2 - 4*16*106)) / (2*16)
Now we can calculate the time for when the ball is 110 feet off the ground by plugging in these values:
t = (85 ± √(7225 - 6784)) / 32
t = (85 ± √4389) / 32
t ≈ (85 ± 66.2) / 32
We get two possible values for time:
t1 ≈ (85 + 66.2) / 32 ≈ 5.91 seconds
t2 ≈ (85 - 66.2) / 32 ≈ 0.91 seconds
Therefore, the ball is 110 feet off the ground at approximately 5.91 seconds and 0.91 seconds after being hit.
h=4+85*time - 1/2 32*time^2
16time^2-85*time-106=0
use the quadratic formula to solve for time (notice two solutions, one going up, one going back down).
time=(85+-sqrt(85^2+4*16*106) )/32
check that, and solve.