An canon ball is launched from ground level directly upward at 39.2 m/s. For how long is the canon ball at or above a height of 34.3 meters?
h = Yo*t + 0.5g*t^2 = 34.3
39.2t - 4.9t^2 = 34.3
-4.9t^2 + 39.2t - 34.3 = 0
t = 1 s. (Used Quad. Formula).
To find the time the cannonball is at or above a height of 34.3 meters, we can use kinematic equations of motion. Specifically, we can use the equation that relates the height, initial velocity, time, and acceleration of the object:
h = vit + (1/2)at^2
Where:
h = height
vi = initial velocity
t = time
a = acceleration
In this case, the cannonball is launched vertically upward, so the initial velocity is positive (+39.2 m/s). The acceleration due to gravity is always negative (-9.8 m/s^2) because it acts downward.
Let's substitute the known values into the equation and solve for time:
34.3 = (39.2)t + (1/2)(-9.8)t^2
Rearranging the equation, we get:
(1/2)(-9.8)t^2 + (39.2)t - 34.3 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, the coefficients are:
a = (1/2)(-9.8)
b = 39.2
c = -34.3
Plugging these values into the quadratic formula, we have:
t = [-(39.2) ± √((39.2)^2 - 4 * (1/2)(-9.8)(-34.3))] / [2 * (1/2)(-9.8)]
Simplifying further:
t = [-39.2 ± √(1536.64 + 671.74)] / -9.8
t = [-39.2 ± √(2208.38)] / -9.8
Calculating the square root and simplifying:
t ≈ [-39.2 ± 46.95] / -9.8
Now we have two possible values for t:
t ≈ (-39.2 + 46.95) / -9.8 ≈ 0.78 seconds
t ≈ (-39.2 - 46.95) / -9.8 ≈ 8.13 seconds
Therefore, the cannonball is at or above a height of 34.3 meters for approximately 0.78 seconds and 8.13 seconds.