A stone is thrown vertically upward with a speed of 18.0 m/s.
(a)How fast is it moving when it reaches a height of 11.0 m?
(b)How long is required to reach this height ?
(c)Why are there two answers to (b)?
first c
It passes 11 meters on the way up and on the way down.
now the calculations
I guess it started at h = 0 (we are lying on our backs in a shallow hole throwing this upward )
Vi = 18
a = -9.81 m/s^2 = g
initial Ke = (1/2) m v^2 = 162 m
loss of Ke at 11 meters = m g h
= 108 m
remaining Ke at 11 meters = (162-108)m = 54 m
so
(1/2) m v^2 = 54 m
v = 10.4 m/s on the way up, -10.4 m/s on the way down by symmetry(part a)
part b
v = Vi - 9.81 t
10.4 = 18 - 9.81 t
t = .775 seconds on the way up
or
-10.4 = 18 - 9.81 t
- 28.4 = -9.81 t
t = 2.9 seconds on the way down
(a) When the stone reaches a height of 11.0 m, it will be moving at a certain speed. To find that speed, we need to analyze the motion of the stone. As the stone is thrown vertically upward, the only force acting on it is the gravitational force. This force causes the stone to slow down as it moves upward, until it reaches its highest point and starts coming back down.
At any point in its trajectory, the velocity of the stone can be calculated using the following equation:
v = u + gt,
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Initially, the stone is moving upward with a speed of 18.0 m/s, so u = 18.0 m/s. The acceleration due to gravity is a constant value of -9.8 m/s^2. Since the stone is moving upward, we'll take the negative sign.
Now, let's find the time taken to reach a height of 11.0 m using the second equation of motion:
s = ut + (1/2)gt^2,
where s is the displacement of the stone. As we want to find the time taken to reach a certain height, we can rearrange the equation as follows:
11.0 = 18.0t + (1/2)(-9.8)t^2.
We can solve the equation to find the time taken. The quadratic equation can be used, and it will give us two solutions - one for when the stone is going upward and another for when it is coming back down.
(b) The reason there are two answers to part (b) is that the stone has to reach a height of 11.0 m twice in its trajectory - once when it is going upward and once when it is coming back down. Therefore, there are two possible times it could take to reach that height. One time corresponds to the stone going upward, and the other time corresponds to the stone coming back down.
So, in short, there are two answers because the stone reaches the given height twice - once on the way up and once on the way down.
To solve the problem, we can use the equations of motion for vertical motion under constant acceleration.
Given:
Initial velocity, u = 18.0 m/s (upward)
Final velocity, v = ?
Initial position, s = 0.0 m
Final position, s = 11.0 m
Acceleration, a = -9.8 m/s² (since the acceleration due to gravity is downward)
(a) To find the velocity when the stone reaches a height of 11.0 m, we can use the equation:
v² = u² + 2as
Substituting the known values:
v² = (18.0 m/s)² + 2(-9.8 m/s²)(11.0 m)
Simplifying:
v² = 324.0 m²/s² - 215.6 m²/s²
v² = 108.4 m²/s²
Taking the square root of both sides:
v = √(108.4 m²/s²)
v ≈ 10.4 m/s
Therefore, the stone is moving at a speed of approximately 10.4 m/s when it reaches a height of 11.0 m.
(b) To find the time required to reach this height, we can use the equation:
v = u + at
Substituting the known values:
10.4 m/s = 18.0 m/s + (-9.8 m/s²)t
Simplifying:
-7.6 m/s = -9.8 m/s²t
Solving for t:
t = (-7.6 m/s) / (-9.8 m/s²)
t ≈ 0.78 s
Therefore, it takes approximately 0.78 seconds to reach a height of 11.0 m.
(c) The reason there are two answers to part (b) is that the stone reaches the 11.0 m height both on its way up and on its way down. The time calculated in part (b) is the time taken to reach the maximum height of the stone. However, the stone continues to move upward for some time and eventually starts coming back down. So, there are two times - one for the upward journey and one for the downward journey.
To answer these questions, we can use the equations of motion for an object in vertical motion. Let's break down each question and explain how to solve them step by step.
(a) How fast is the stone moving when it reaches a height of 11.0 m?
To determine the speed of the stone at 11.0 m height, we need to consider its initial velocity and the acceleration due to gravity. Here's how we can find the answer:
1. Identify the data given:
- Initial velocity (u) = 18.0 m/s
- Height (s) = 11.0 m
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming negligible air resistance)
2. Determine the final velocity (v) using the equation:
v^2 = u^2 + 2as
Substitute the values:
v^2 = (18.0 m/s)^2 + 2(-9.8 m/s^2)(11.0 m)
v^2 = 324 m^2/s^2 - 215.6 m^2/s^2
v^2 = 108.4 m^2/s^2
Take the square root to find the velocity:
v ≈ √108.4 m^2/s^2
v ≈ 10.414 m/s (rounded to three decimal places)
Therefore, the stone is moving at approximately 10.414 m/s when it reaches a height of 11.0 m.
(b) How long is required to reach this height?
To find the time taken by the stone to reach a height of 11.0 m, we can use the equation for vertical displacement as a function of time. Here's how to solve it:
1. Identify the data given:
- Initial velocity (u) = 18.0 m/s
- Height (s) = 11.0 m
- Acceleration due to gravity (g) = 9.8 m/s^2
2. Determine the time (t) using the equation:
s = ut + (1/2)gt^2
Substitute the values:
11.0 m = (18.0 m/s)t + (1/2)(9.8 m/s^2)t^2
Rearrange the equation to form a quadratic equation:
(1/2)(9.8 m/s^2)t^2 + (18.0 m/s)t - 11.0 m = 0
3. Solve the quadratic equation either by factoring, completing the square, or using the quadratic formula.
In this case, using the quadratic formula is the most straightforward method:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation (ax^2 + bx + c = 0), it becomes:
t = [-(18.0 m/s) ± √((18.0 m/s)^2 - 4(1/2)(9.8 m/s^2)(-11.0 m))] / [2(1/2)(9.8 m/s^2)]
4. Calculate the two possible values for t using the formula:
t1 = [-(18.0 m/s) + √((18.0 m/s)^2 - 4(1/2)(9.8 m/s^2)(-11.0 m))] / [2(1/2)(9.8 m/s^2)]
t2 = [-(18.0 m/s) - √((18.0 m/s)^2 - 4(1/2)(9.8 m/s^2)(-11.0 m))] / [2(1/2)(9.8 m/s^2)]
(c) Why are there two answers to (b)?
In this case, there are two solutions to the quadratic equation, resulting in two possible values for time (t1 and t2).
One value represents the time it takes for the stone to reach a height of 11.0 m on its way up (ascending), and the other value represents the time it takes for the stone to reach the same height on its way down (descending) after reaching its maximum height.
Since the stone is thrown upwards, the positive value of 't' (t1) corresponds to the time it takes for the stone to reach 11.0 m while ascending. The negative value of 't' (t2) corresponds to the time it takes for the stone to reach 11.0 m while descending.
Therefore, there are two answers to question (b), representing the ascending and descending times of the stone at a height of 11.0 m.