An attacker at the base of a castle wall 3.82 m high throws a rock straight up with speed 7.10 m/s from a height of 1.45 m above the ground.
(a) Will the rock reach the top of the wall?
__yes
__no
(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?
__ m/s
(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 7.10 m/s and moving between the same two points.
__ m/s
none
a) yes
To solve this problem, we can use the kinematic equations of motion.
(a) To determine if the rock will reach the top of the wall, we can compare the maximum height it can achieve with the height of the wall.
Let's calculate the maximum height reached by the rock first.
Using the equation:
vf^2 = vi^2 + 2ad
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the displacement.
Since the rock is thrown straight up, the final velocity at the maximum height will be 0 m/s. The acceleration due to gravity (g) will act in the opposite direction, causing the rock to decelerate.
Thus, we have:
0^2 = (7.10 m/s)^2 - 2 * g * d
Simplifying:
0 = 50.41 m^2/s^2 - 2 * 9.8 m/s^2 * d
Rearranging to solve for d:
2 * 9.8 m/s^2 * d = 50.41 m^2/s^2
d = 50.41 m^2/s^2 / (2 * 9.8 m/s^2)
d ≈ 2.58 m
From the calculations, we find that at its maximum height, the rock reaches approximately 2.58 m.
Comparing this value to the height of the wall (3.82 m), we can conclude that the rock will not reach the top of the wall.
__no
(b) Since the rock will not reach the top of the wall, we need to calculate the initial speed it must have to reach that height.
Using the same kinematic equation:
vf^2 = vi^2 + 2ad
We can substitute the values:
0^2 = vi^2 + 2 * -9.8 m/s^2 * (3.82 m - 1.45 m)
Simplifying:
0 = vi^2 - 2 * 9.8 m/s^2 * 2.37 m
Rearranging to solve for vi:
vi^2 = 2 * 9.8 m/s^2 * 2.37 m
vi ≈ √(2 * 9.8 m/s^2 * 2.37 m)
vi ≈ √(46.3 m^2/s^2)
vi ≈ 6.8 m/s
Hence, the rock must have an initial speed of approximately 6.8 m/s to reach the top of the wall.
__ 6.8 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 formula:
vf^2 = vi^2 + 2ad
In this case, the rock is thrown downward with an initial speed of 7.10 m/s. The final velocity at the height of 1.45 m will be 0 m/s, as the rock comes to a stop.
Plugging in the values:
0^2 = (7.10 m/s)^2 + 2 * -9.8 m/s^2 * (1.45 m - 3.82 m)
Simplifying:
0 = 50.41 m^2/s^2 + 2 * -9.8 m/s^2 * (-2.37 m)
Rearranging to solve for the change in speed:
Change in speed = √(50.41 m^2/s^2 + 2 * 9.8 m/s^2 * 2.37 m)
Change in speed ≈ √(66.0534 m^2/s^2)
Change in speed ≈ 8.12 m/s
Therefore, the change in speed of the rock thrown straight down between the two points is approximately 8.12 m/s.
__ 8.12 m/s
To solve these physics problems, we can use the equations of motion and the principles of kinematics.
(a) To determine whether the rock will reach the top of the wall, we need to compare its height at its highest point and the height of the wall.
At the highest point of the rock's trajectory, its height above the ground is given by the equation:
h = h0 + V0y * t - (1/2) * g * t^2,
where h is the height, h0 is the initial height, V0y is the initial vertical velocity, t is time, and g is the acceleration due to gravity.
Substituting the known values, we have:
h = 3.82 m,
h0 = 1.45 m,
V0y = 7.10 m/s,
g = 9.8 m/s^2.
We can rearrange the equation to solve for the time it takes for the rock to reach its highest point:
t = (V0y ± √(V0y^2 - 2g(h - h0))) / g,
where the ± sign accounts for the fact that the rock can either reach the highest point or fall back down.
Calculating the value inside the square root:
V0y^2 - 2g(h - h0) = 7.10^2 - 2 * 9.8 * (3.82 - 1.45) = -0.0808.
Since we have a negative value inside the square root, we will not be able to take the square root of a negative number.
Therefore, the rock will not reach the top of the wall.
Answer: no.
(b) Since the rock does not reach the top of the wall, we need to calculate the initial speed required for it to reach the top.
Using the same equation as before:
V0y = √(V0y^2 - 2g(h - h0)),
we can solve for V0y.
Substituting the known values:
V0y = √(7.10^2 - 2 * 9.8 * (3.82 - 1.45)) = 9.97 m/s.
Answer: 9.97 m/s.
(c) To find the change in speed of a rock thrown straight down from the top of the wall, we can use a similar approach.
At the highest point of the rock's trajectory, its speed is given by the equation:
V = √(V0^2 + 2g(h - h0)),
where V is the speed, V0 is the initial speed, h is the height, h0 is the initial height, and g is the acceleration due to gravity.
Substituting the known values:
V0 = -7.10 m/s (negative because the rock is thrown downward),
h = 0 m (starting and ending at ground level),
h0 = 3.82 m,
g = 9.8 m/s^2,
V = √((-7.10)^2 + 2 * 9.8 * (0 - 3.82)) = 19.0 m/s.
The initial speed of the rock thrown straight down is 7.10 m/s, so the change in speed is:
Change in speed = V - V0 = 19.0 m/s - 7.10 m/s = 11.9 m/s.
Answer: 11.9 m/s.