A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Take the acceleration due to gravity to have magnitude and neglect any effects of air resistance. With what speed was the rock thrown?
Some people keep suggesting to use the equation:
y = yo + vot + 1/2at^2
set up like: 0 = 56.3 + vo(4) + 1/2(9.8)(4^2)
But that can't be right, because 4 seconds is the time it took from when it left the throwers hand to when it hit the ground. Not the time it took to travel 56.3 meters.
A train 900'm long takes 1minute to cover a bridge of 300 m calculate its average speed
You are correct. The equation you mentioned, y = yo + vot + 1/2at^2, is not directly applicable in this case because it assumes constant acceleration throughout the motion. However, when an object is thrown upwards, its acceleration changes sign from positive to negative as it reaches its maximum height and begins to fall downward.
To solve this problem, we can divide the motion into two separate parts: the upward motion and the downward motion. Let's look at them individually:
1. Upward motion: During the upward motion, the initial velocity (vo) is positive, and the acceleration (a) is equal to the magnitude of the acceleration due to gravity, but in the opposite direction (let's take it as -9.8 m/s^2 to indicate downward direction). The final velocity (v) is zero when the rock reaches its maximum height. We want to find the initial velocity (vo).
Using the equation v = vo + at, we can plug in the known values:
0 = vo + (-9.8)(t1)
vo = 9.8(t1)
2. Downward motion: During the downward motion, the initial velocity is now negative because the rock is moving downward, and the acceleration remains -9.8 m/s^2. The final velocity will be the velocity at impact when the rock hits the ground, and we want to find its magnitude (|v|).
Using the equation y = yo + vot + 1/2at^2, where yo is the initial position, y is the final position, vo is the initial velocity, t is the time taken, and a is the acceleration, we can rearrange the equation to solve for |v|:
y = yo + vot + 1/2at^2
-56.3 = 0 + (-9.8)(t2) + 1/2(-9.8)(t2)^2
Simplifying, we get:
-56.3 = -4.9(t2)^2
Now, you mentioned that it took 4 seconds for the rock to hit the ground. Let's call the time taken for the upward motion as t1 and the time taken for the downward motion as t2. The total time taken will be equal to t1 + t2, which is 4 seconds.
Therefore, t2 = 4 - t1.
Substituting this back into the simplified equation, we get:
-56.3 = -4.9(4 - t1)^2
Solving for t1 will give us the time taken for the upward motion.
With the value of t1, we can then use vo = 9.8(t1) to find the initial velocity when the rock is thrown upwards.
Please provide the value of t1 so that we can proceed with the calculations and determine the speed at which the rock was thrown.
You are correct in your analysis. The equation that has been suggested, y = yo + vot + 1/2at^2, is not the appropriate equation to solve this problem because it assumes constant acceleration and does not account for the fact that the rock was thrown vertically upward and then came back down.
To solve this problem, we need to break it down into two parts: the upward motion and the downward motion of the rock.
First, let's consider the upward motion. Here are the known values for the upward motion:
Initial velocity (vo) = ?
Acceleration (a) = -9.8 m/s^2 (negative because it acts in the opposite direction of the thrown motion)
Time of flight (t) = ? (this is what we are trying to find)
Displacement (y) = 56.3 meters (the height of the building)
We can use the equation: y = yo + vot + 1/2at^2, but in this case, yo is not 0 because the rock was thrown from a height of 0. Therefore, the equation becomes: y = yo + vot + 1/2at^2
Plugging in the values we have:
56.3 = 0 + vo(t) + 1/2(-9.8)(t^2)
Now let's consider the downward motion. The rock is falling from a height of 56.3 meters to the ground (height = 0). We can again use the same equation, but this time, our displacement (y) is 0.
0 = 56.3 + 0(t) + 1/2(-9.8)(t^2)
At this point, we have two equations with two unknowns (vo and t). We can solve this system of equations simultaneously to find the values.
Alternatively, we can solve this problem using energy considerations. At the top of its motion, the rock has only potential energy. At the bottom of its motion, it has only kinetic energy. We can equate the two and solve for vo.
Potential energy at the top = Kinetic energy at the bottom
mgh = (1/2)mv^2
We cancel out the mass (m) on both sides of the equation, and the height (h) is 56.3 meters.
56.3(9.8) = (1/2)v^2
Simplifying further, we get:
v^2 = 56.3(9.8) * 2
v^2 = 1091.72
v ≈ 33.03 m/s
Therefore, the rock was thrown with a speed of approximately 33.03 m/s.