An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.50 m above the ground.
(a) Will the rock reach the top of the wall?
(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?
(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5.00 m/s and moving between the same two points.
(d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why it does or does not agree.
To solve this problem, we can use the principles of projectile motion. Let's break it down into parts:
(a) To determine whether the rock will reach the top of the wall, we can use the equations of motion. Here, the initial height is 1.50 m, and the final height is 3.95 m. The acceleration due to gravity, which acts in the downward direction, is -9.8 m/s^2. We can use the equation:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * displacement
Since the rock is thrown straight up from the base of the wall, the displacement is the difference between the final height and the initial height. Plugging in the values, we get:
Final velocity^2 = 5.00 m/s^2 + 2 * (-9.8 m/s^2) * (3.95 m - 1.50 m)
Simplifying the equation, we find:
Final velocity^2 = 5.00 m^2/s^2 - 2 * 9.8 m/s^2 * 2.45 m
Final velocity^2 = 5.00 m^2/s^2 - 48.02 m^2/s^2
Final velocity^2 ≈ -43.02 m^2/s^2
Since velocity cannot be negative in this context, the rock will not reach the top of the wall.
(b) Since the rock does not reach the top, the second part of the question asks for the initial speed required to reach the top of the wall. To find this speed, we can use the same equation mentioned in part (a), but with the displacement equal to the total height of the wall (3.95 m) and the final velocity equal to zero (since it reaches the top). Plugging in the values, we obtain:
0 = Initial velocity^2 + 2 * (-9.8 m/s^2) * (3.95 m)
Simplifying further, we have:
0 = Initial velocity^2 - 2 * 9.8 m/s^2 * 3.95 m
Initial velocity^2 = 2 * 9.8 m/s^2 * 3.95 m
Initial velocity^2 = 77.46 m^2/s^2
Taking the square root of both sides, we get:
Initial velocity ≈ 8.80 m/s
Therefore, the rock needs an initial speed of approximately 8.80 m/s to reach the top of the wall.
(c) To find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5.00 m/s, we can use a similar approach. In this case, the initial velocity is 5.00 m/s, the final velocity is 0 m/s (as it reaches the ground), and the displacement is the same as the height of the wall (3.95 m). Using the equation:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * displacement
0 = 5.00 m/s^2 + 2 * 9.8 m/s^2 * 3.95 m
0 = 5.00 m/s^2 + 2 * 9.8 m/s^2 * 3.95 m
Simplifying the equation, we find:
Final velocity^2 = 5.00 m/s^2 + 77.46 m/s^2
Final velocity^2 ≈ 82.46 m/s^2
Taking the square root of both sides, we get:
Final velocity ≈ 9.09 m/s
Therefore, the change in speed of the rock thrown straight down from the top of the wall is approximately 9.09 m/s.
(d) The change in speed of the downward-moving rock does not agree with the magnitude of the speed change of the rock moving upward. This is because the rock moving downward experiences the acceleration due to gravity (9.8 m/s^2) acting in the same direction as its velocity, while the rock moving upward experiences the acceleration due to gravity acting in the opposite direction to its velocity. As a result, the magnitude of the change in speed for the upward-moving rock is greater than the change in speed for the downward-moving rock, assuming the same elevation change.
To answer these questions, we can use the principles of projectile motion. Let's break down each part step-by-step:
(a) To determine if the rock will reach the top of the wall, we need to find its maximum height. We can use the equation for the vertical motion of a projectile:
v_f^2 = v_i^2 - 2gΔy
where v_f is the final velocity (which will be 0 at the top), v_i is the initial velocity, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and Δy is the change in height.
Given:
v_i = 5.00 m/s (upwards)
Δy = 3.95 m - 1.50 m = 2.45 m
Plugging these values into the equation, we can solve for v_f:
0 = (5.00 m/s)^2 - 2 * 9.8 m/s^2 * 2.45 m
Solving this equation, we find that the final velocity v_f = 12.13 m/s (upwards).
Since the final velocity is positive, the rock will indeed reach the top of the wall.
(b) To find the speed of the rock at the top of the wall, we can use the equation for velocity:
v_f = v_i - gt
Since the rock reaches its maximum height at the top of the wall, the time it takes to reach the top is half of the total time of flight.
The time of flight can be found using the equation:
Δy = v_i * t - 0.5 * g * t^2
where Δy is the vertical distance traveled.
Substituting the values:
2.45 m = 5.00 m/s * t - 0.5 * 9.8 m/s^2 * t^2
Solving this quadratic equation, we find two possible solutions: t ≈ 0.6528 s and t ≈ 0.0000 s. The second solution is not physically meaningful since it corresponds to the initial position.
Therefore, the time of flight is approximately 0.6528 s.
Now, let's find the velocity at the top using the equation:
v_f = v_i - gt
v_f = 5.00 m/s - 9.8 m/s^2 * 0.6528 s ≈ -2.15 m/s
The absolute value of the velocity gives us the speed at the top, which is approximately 2.15 m/s.
(c) To find the change in speed of a rock thrown straight down from the top of the wall, we can use the same equation we used in part (b) for the final downward velocity:
v_f = v_i - gt
Given:
v_i = 5.00 m/s (downwards)
We can find the final velocity v_f when the rock hits the ground by assuming Δy = 3.95 m:
0 = 5.00 m/s - 9.8 m/s^2 * t
Solving for t, we find t ≈ 0.51 s.
The change in speed will be the final velocity since the initial velocity is zero:
Δv = v_f = -9.8 m/s^2 * 0.51 s ≈ -5.00 m/s
The change in speed is approximately -5.00 m/s.
(d) Since the magnitude of the change in speed of the downward-moving rock is 5.00 m/s and the magnitude of the change in speed of the upward-moving rock is 2.15 m/s, we can see that they do not agree.
This difference in magnitude is due to the fact that the upward-moving rock experiences a greater deceleration (due to gravity) over a longer time compared to the downward-moving rock. As a result, the upward-moving rock has a smaller change in speed compared to the downward-moving rock.
v = 5 - 9.8 t
0 = 5 - 9.8 t
t = 5/9.8 = .51 seconds to top where v = 0
average speed going up = 5/2 = 2.5 m/s
h = 1.5 + 2.5 * .51 = 2.78 meters
(a) no way
(b)The same speed it will have if dropped from the top of the wall and landing on a platform 1.5 m high
3.95 - 1.5 = 2.45 meters down
2.45 = (1/2)g t^2
2.45 = 4.9 t^2
t = .707 seconds
v = g t = 9.8*.707 = 6.93 meters/second initial speed up
c and d
change in speed /change in time = g