A rock is thrown straight up at 22.0 m/s from a roof of a 30.0 m tall building. After rising, the rock falls on the street below. Ignoring air resistance, how much time is the rock in the air?
Show the work please :)
hf=hi+vi*t-4.8t^2
4.8t^2-22t-30=0
use the quadratic equation to solve for time t.
5.59 seconds
To solve this problem, we can use the kinematic equation for vertical motion:
y = y0 + v0*t + (1/2)*a*t^2
where:
- y is the vertical position at time t
- y0 is the initial vertical position (30.0 m)
- v0 is the initial vertical velocity (22.0 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
Since the rock is thrown straight up, its final vertical position is zero when it falls back down to the ground. We can set y = 0 and solve for t.
0 = 30.0 + 22.0*t + (1/2)*(-9.8)*t^2
Simplifying the equation, we get:
4.9*t^2 + 22.0*t - 30.0 = 0
Now we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = 4.9, b = 22.0, and c = -30.0. Let's substitute these values and calculate the values of t:
t = (-22.0 ± √(22.0^2 - 4*4.9*(-30.0))) / (2*4.9)
t = (-22.0 ± √(484 + 588)) / 9.8
t = (-22.0 ± √1072) / 9.8
Using a calculator, we find:
t ≈ -0.627 seconds (not valid since time cannot be negative)
t ≈ 2.75 seconds
Therefore, the rock is in the air for approximately 2.75 seconds.
To find the time the rock is in the air, we can use the equation of motion for vertical motion under constant acceleration.
First, let's determine the time it takes for the rock to reach its maximum height. We can use the formula:
time = (final velocity - initial velocity) / acceleration
In this case, the initial velocity is 22.0 m/s, the final velocity is 0 m/s when the rock reaches its maximum height, and the acceleration is -9.8 m/s² (due to gravity and assuming upward as positive).
Using the formula, we have:
time_up = (0 - 22.0) / -9.8
time_up = 2.24 seconds
Next, we need to find the time it takes for the rock to fall from the maximum height to the ground. The distance the rock falls is equal to the height of the building, which is 30.0 meters.
We can use the equation:
distance = initial velocity * time + 0.5 * acceleration * time²
Since the initial velocity is 0 m/s when the rock starts falling, we have:
distance = 0.5 * acceleration * time_down²
Plugging in the known values:
30.0 = 0.5 * -9.8 * time_down²
Simplifying the equation:
30.0 = -4.9 * time_down²
Dividing both sides by -4.9:
time_down² = -30.0 / -4.9
time_down² = 6.12
Taking the square root of both sides:
time_down = √6.12
time_down ≈ 2.47 seconds
Finally, to calculate the total time the rock is in the air, we sum the time for upward motion and the time for downward motion:
total time = time_up + time_down
total time = 2.24 + 2.47
total time ≈ 4.71 seconds
Therefore, the rock is in the air for approximately 4.71 seconds.