Suppose you throw a stone straight up with an initial speed of 15 m s−1
.
(a) (5 points) If you throw a second stone straight up 1.0 s after the first, with what speed must you throw
the second stone if it is to hit the first at a height of 11.0 m? There are two answers. Do both solutions
make physical sense?
(b) (2 points) Plot x − t and v − t graphs for both balls.
(c) (3 points) If you throw the second stone 1.30 s after the first, with what speed must you throw the second
stone if it is to hit the first at a height of 11.0 m?
(a) when is the first stone at 11m?
15t-4.9t^2 = 11
t = 1.21, 1.84
So, you need the 2nd stone to be at 11m one second faster.
v(t-1)-4.9(t-1)^2 = 11
v = 53.41
Repeat with t=1.84 to meet on the way down.
(b) I assume you can graph parabolas
(c) change t-1 to t-1.3
(a) Well, if you throw the second stone 1.0 s after the first stone, they both have the same time of flight. So, in order for the second stone to hit the first stone at a height of 11.0 m, it must have the same final velocity as the first stone when it reaches that height. Therefore, to find the required velocity of the second stone, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Since the acceleration is due to gravity and is a constant value of -9.8 m/s^2, the equation becomes v = u - 9.8t.
Now, let's calculate the velocity.
For the first stone, v1 = 0 m/s (since it reaches its maximum height and then comes back down), u1 = 15 m/s, and t = 1.0 s. Plugging these values into the equation, we get 0 = 15 - 9.8(1.0), which gives us v1 = 15 + 9.8 = 24.8 m/s.
For the second stone, v2 = 0 m/s (since it reaches its maximum height and then comes back down), t = 1.0 s, and since it starts 1.0 s later than the first stone, the time taken for it is 1.0 s. Plugging these values into the equation, we get 0 = u2 - 9.8(1.0), which gives us u2 = 9.8 m/s.
So, there are two possible solutions:
1. If you throw the second stone with an initial speed of 24.8 m/s, it will hit the first stone at a height of 11.0 m.
2. If you throw the second stone with an initial speed of 9.8 m/s, it will hit the first stone at a height of 11.0 m.
Both of these solutions make physical sense because they satisfy the given conditions.
(b) Since the two stones have the same time of flight, the x - t and v - t graphs for both stones will be symmetrical. The x - t graph will be a parabolic curve (an inverted U-shape), while the v - t graph will be a straight line sloping downward.
(c) If you throw the second stone 1.30 s after the first, the time taken for the first stone will be 1.30 s and the time taken for the second stone will be 1.30 s - 1.0 s = 0.30 s. Using the same equation v = u - 9.8t, we can calculate the required velocity for the second stone.
For the first stone, v1 = 0 m/s, u1 = 15 m/s, and t = 1.30 s. Plugging these values into the equation, we get 0 = 15 - 9.8(1.30), which gives us v1 = 15 - 12.74 = 2.26 m/s.
For the second stone, v2 = 0 m/s, t = 0.30 s, and u2 is what we need to find. So, the equation becomes 0 = u2 - 9.8(0.30), which gives us u2 = 2.94 m/s.
Therefore, to hit the first stone at a height of 11.0 m, you need to throw the second stone with an initial speed of either 2.26 m/s or 2.94 m/s.
(a) To find the speed at which the second stone must be thrown in order to hit the first stone at a height of 11.0 m, we need to use the equations of motion.
Let's start by finding the time it takes for the first stone to reach a height of 11.0 m. We'll use the equation:
h = ut + 0.5gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Since the stone is thrown straight up, the initial velocity (u) is positive 15 m/s, and the acceleration due to gravity (g) is negative 9.8 m/s^2 (taking downward as the positive direction).
11.0 = 15t - 0.5(9.8)t^2
Rearranging the equation:
0.5(9.8)t^2 - 15t + 11 = 0
Now we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 0.5(9.8), b = -15, and c = 11.
Solving for t, we find two solutions: t ≈ 0.79 s and t ≈ 2.13 s.
Now we need to find the speed at which the second stone must be thrown in order to hit the first stone at those times.
For t = 0.79 s:
Let's assume the second stone is thrown with an initial velocity v. The equation for the height of the second stone is:
h = vt - 0.5gt^2
Substituting the values:
11.0 = v(0.79) - 0.5(9.8)(0.79)^2
Simplifying the equation:
11.0 = 0.79v - 3.91
0.79v = 14.91
v ≈ 18.9 m/s
For t = 2.13 s:
11.0 = v(2.13) - 0.5(9.8)(2.13)^2
11.0 = 2.13v - 23.63
2.13v = 34.63
v ≈ 16.3 m/s
So, the two speeds at which the second stone must be thrown are approximately 18.9 m/s and 16.3 m/s.
Both solutions make physical sense because the first stone is moving upwards with a decreasing velocity due to gravity. The second stone needs to be thrown with a higher speed to catch up with the first stone when it reaches a height of 11.0 m.
(b) To plot the x−t and v−t graphs for both balls, we first need to find the position and velocity equations for each ball.
For the first ball:
x1 = ut + 0.5gt^2
v1 = u - gt
For the second ball:
x2 = v2t - 0.5gt^2
v2 = v2 - gt
Now we can plot the graphs:
x−t graph:
- For the first ball, the position (x1) will increase until it reaches its maximum height and then decrease.
- For the second ball, the position (x2) will start from 0 and increase until it reaches the height of the first ball and then decrease.
v−t graph:
- For the first ball, the velocity (v1) will decrease uniformly due to gravity until it reaches 0 at its maximum height and then increase in the downward direction.
- For the second ball, the velocity (v2) will increase uniformly due to gravity in the downward direction.
(c) To find the speed at which the second stone must be thrown in order to hit the first stone at a height of 11.0 m after 1.30 s, we follow the same steps as in part (a) but with a different time.
Using the equation:
11.0 = v(1.30) - 0.5(9.8)(1.30)^2
Solving for v:
v ≈ 14.3 m/s
Therefore, the speed at which the second stone must be thrown is approximately 14.3 m/s.
To solve this problem, we can use the equations of motion for an object in free fall. Let's break down the steps for each part of the question:
(a) To find the initial speed of the second stone so that it hits the first stone at a height of 11.0 m, we need to find when and where the first stone will be at that height.
1. Find the time it takes for the first stone to reach a height of 11.0 m:
We can use the equation h = ut + (1/2)gt^2, where h is the height, u is the initial speed, g is the acceleration due to gravity (approximated as -9.81 m/s^2), and t is the time.
Plugging in the given values, we get 11.0 = 15t - (1/2)(9.81)t^2.
2. Solve the equation for t:
Rearrange the equation to get (1/2)(9.81)t^2 - 15t + 11.0 = 0. Solve this quadratic equation to find the values of t.
3. Once we have the time(s) at which the first stone is at a height of 11.0 m, we can calculate the speed of the second stone using the formula v = u + gt, where u is the initial speed and g is the acceleration due to gravity.
(b) To plot the x - t and v - t graphs for both balls, we need to use the equations of motion:
1. For the x - t graph, the position of an object in free fall can be determined using the equation x = ut + (1/2)gt^2, where x is the position, u is the initial speed, g is the acceleration due to gravity, and t is the time. Plot x on the y-axis and t on the x-axis for both stones, using the calculated time values.
2. For the v - t graph, the velocity of an object in free fall can be determined by differentiating the equation x = ut + (1/2)gt^2 with respect to time. The result is v = u + gt, where v is the velocity, u is the initial speed, g is the acceleration due to gravity, and t is the time. Plot v on the y-axis and t on the x-axis for both stones, using the calculated time values.
(c) Repeat steps 1 and 2 from part (a) with a time delay of 1.30 s instead of 1.0 s.
Remember to check that the solutions make physical sense. For example, if the calculated initial speed of the second stone is negative, it means it should be thrown downward rather than upward. Also, consider the practical limitations and assumptions of the problem.