A brick is thrown vertically upward with an initial speed of 4.40 m/s from the roof of a building. If the building is 74.0 m tall, how much time passes before the brick lands on the ground?
Solve this equation for t:
-74 = 4.4 t - (g/2) t^2
g = 9.8 m/s^2
Take the root that is positive.
To find the time it takes for the brick to land on the ground, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
- s is the distance traveled (74.0 m, the height of the building)
- u is the initial velocity (4.40 m/s, the speed at which the brick is thrown upwards)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming downward acceleration)
- t is the time
As the brick is thrown upward, the acceleration due to gravity should be negative in this case.
Let's substitute the values into the equation:
74.0 = (4.40)t + (1/2)(-9.8)t^2
Now, we can rearrange the equation to solve for time:
(1/2)(-9.8)t^2 + (4.40)t - 74.0 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = (1/2)(-9.8) = -4.9, b = 4.40, and c = -74.0.
t = (-4.40 ± sqrt(4.40^2 - 4(-4.9)(-74.0))) / (2(-4.9))
Now we can calculate the values inside the square root:
t = (-4.40 ± sqrt(19.36 + 1443.2)) / (-9.8)
t = (-4.40 ± sqrt(1462.56)) / (-9.8)
t = (-4.40 ± 38.22) / (-9.8)
We have two possible solutions:
t1 = (-4.40 + 38.22) / (-9.8)
t2 = (-4.40 - 38.22) / (-9.8)
Calculating these values:
t1 = 33.82 / (-9.8)
t2 = -42.62 / (-9.8)
t1 ≈ -3.45 seconds (not a valid solution)
t2 ≈ 4.35 seconds
Since time cannot be negative, the valid solution is t = 4.35 seconds.
Therefore, it will take approximately 4.35 seconds for the brick to land on the ground.