A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11 m/s when the hand is 1.8 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
hf=hi+ vi*t-4.9t^2
0=1.8+11t-4.9t^2
solve the quadratic equation for t.
2.39
3.7
To find the time it takes for the ball to hit the ground, we need to use the equation of motion for the vertical direction. The equation is expressed as:
h = h0 + v0t - (1/2)gt^2
Where:
h = final height (in this case, 0 since it hits the ground)
h0 = initial height (1.8 m)
v0 = initial velocity (11 m/s)
g = acceleration due to gravity (assumed to be 9.8 m/s^2)
t = time
By substituting the known values into the equation, we can solve for t:
0 = 1.8 + 11t - (1/2)(9.8)t^2
To simplify the equation, let's multiply everything by 2 to get rid of the fraction:
0 = 3.6 + 22t - 9.8t^2
Rearranging the equation to form a quadratic equation:
9.8t^2 - 22t - 3.6 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
By substituting the values into the formula:
t = (-(-22) ± √((-22)^2 - 4 * 9.8 * -3.6)) / (2 * 9.8)
Simplifying further:
t = (22 ± √(484 + 140.16)) / 19.6
t = (22 ± √(624.16)) / 19.6
t = (22 ± 24.98) / 19.6
Now, we have two possible solutions:
t1 = (22 + 24.98) / 19.6 ≈ 2.55 seconds
t2 = (22 - 24.98) / 19.6 ≈ -0.13 seconds
Since we are only interested in the positive value for time, the ball is in the air for approximately 2.55 seconds before it hits the ground.