A ball it hit straight up in the air with an initial velocity of 12.9 m/s. At what time does it cross the height of 5 m going up?
5 = 12.9 t - 4.9 t^2
4.9 t^2 - 12.9 t + 5 = 0
t = 0.472 going up and 2.16 coming down
To find the time at which the ball crosses the height of 5 m going up, we can use the equations of motion for vertical motion.
The equation that relates the displacement, initial velocity, time, and acceleration for vertical motion is:
s = ut + (1/2)at²
Where:
s = displacement (change in height)
u = initial velocity
t = time
a = acceleration (which is acceleration due to gravity, approximately -9.8 m/s²)
In this case, we know:
u = 12.9 m/s (upward velocity)
s = 5 m (height)
Since the ball is going up, its initial velocity is in the opposite direction to the acceleration due to gravity. Therefore, we can substitute the negative value for acceleration (-9.8 m/s²) in the equation.
Using this equation, we can rearrange it to solve for time (t):
s = ut + (1/2)at²
5 = (12.9)t + (1/2)(-9.8)t²
Simplifying the equation further, we have:
5 = 12.9t - 4.9t²
Rearranging the equation to a quadratic form:
4.9t² - 12.9t + 5 = 0
Now, we can solve this quadratic equation 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 ± √(b² - 4ac)) / (2a)
For our equation:
a = 4.9
b = -12.9
c = 5
Plugging in the values, we get:
t = (-(-12.9) ± √((-12.9)² - 4(4.9)(5))) / (2 * 4.9)
Simplifying it further, we have:
t = (12.9 ± √(166.41 - 98)) / 9.8
t = (12.9 ± √(68.41)) / 9.8
t = (12.9 ± 8.27) / 9.8
Now, we have two possible solutions for time:
t₁ = (12.9 + 8.27) / 9.8
t₂ = (12.9 - 8.27) / 9.8
Evaluating the values, we get:
t₁ ≈ 2.18 seconds
t₂ ≈ 0.48 seconds
Therefore, the ball crosses the height of 5 m going up at approximately 2.18 seconds and 0.48 seconds.
To find the time it takes for the ball to reach a height of 5 m going up, we can use the kinematic equation for vertical motion:
h = v₀t + (1/2)at²
Where:
h = height (5 m)
v₀ = initial velocity (12.9 m/s)
t = time (unknown)
a = acceleration due to gravity (-9.8 m/s²)
Rearranging the equation, we get:
5 = 12.9t + (1/2)(-9.8)t²
Simplifying the equation, we have:
-4.9t² + 12.9t - 5 = 0
We can solve this equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values, we have:
t = (-(12.9) ± √((12.9)² - 4(-4.9)(-5))) / (2(-4.9))
Simplifying further:
t = (-12.9 ± √(166.41 - 98)) / (-9.8)
t = (-12.9 ± √(68.41)) / (-9.8)
t = (-12.9 ± 8.27) / (-9.8)
Now, we have two possible solutions for t:
t1 = (-12.9 + 8.27) / (-9.8) ≈ 0.448 s
t2 = (-12.9 - 8.27) / (-9.8) ≈ 2.056 s
Since the ball is going up, we only consider the positive solution, which is approximately 0.448 seconds.
Therefore, it takes approximately 0.448 seconds for the ball to cross the height of 5 m going up.