During a hockey game the puck gets flipped up from the ice, over the boards, and
lands on the head of an unsuspecting fan 7.5 m above ice level. The puck was shot
up at an angle of 65° and was in the air for 1.8 s.
(a) Calculate the vertical component of the puck's initial velocity. (Ans: 13 m/s [up])
(b) Calculate the puck's initial velocity. (Ans: 14.3 m/s [65° above horizon])
(c) Determine the maximum height of the puck above ice level. (Ans: 8.6 m)
(a) To find the vertical component of the puck's initial velocity, we can use the equation:
v = v0 + at
where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, since the puck is shot straight up and comes back down, the final velocity will be zero, so we can rewrite the equation as:
0 = v0 + (-9.8 m/s^2)(1.8 s)
Solving for v0, we get:
v0 = (9.8 m/s^2)(1.8 s)
v0 = 17.64 m/s
The vertical component of the puck's initial velocity is therefore 17.64 m/s downwards. However, since the question asks for the vertical component, we need to reverse the direction and take the absolute value:
Vertical component of the puck's initial velocity = |-17.64 m/s| = 17.64 m/s [up]
Therefore, the vertical component of the puck's initial velocity is 17.64 m/s [up].
(b) To find the puck's initial velocity, we can use the equation:
v = √(v_x^2 + v_y^2),
where v_x is the horizontal component of the velocity and v_y is the vertical component of the velocity. We already found the vertical component of the velocity in part (a), which was 17.64 m/s [up]. To find the horizontal component, we can use:
v_x = v * cos(theta),
where theta is the angle at which the puck was shot up. In this case, theta = 65°.
v_x = 17.64 m/s * cos(65°)
v_x = 17.64 m/s * 0.4226
v_x = 7.46 m/s
Now that we have the vertical and horizontal components of the velocity, we can use the equation above to find the initial velocity:
v = √(v_x^2 + v_y^2)
v = √((7.46 m/s)^2 + (17.64 m/s)^2)
v = √(55.5316 m^2/s^2 + 310.3296 m^2/s^2)
v = √365.8612 m^2/s^2
v ≈ 19.1 m/s
Therefore, the puck's initial velocity is approximately 19.1 m/s [65° above horizon].
(c) To find the maximum height of the puck above ice level, we can use the equation:
h = v_y * t + (1/2) * a * t^2,
where h is the maximum height, v_y is the vertical component of the velocity, t is the time, and a is the acceleration due to gravity (9.8 m/s^2).
In this case, we already know that v_y = 17.64 m/s and t = 1.8 s. Plugging these values into the equation, we get:
h = (17.64 m/s)(1.8 s) + (1/2)(9.8 m/s^2)(1.8 s)^2
h = 31.752 m + (1/2)(9.8 m/s^2)(3.24 s^2)
h = 31.752 m + (1/2)(9.8 m/s^2)(10.4976 s^2)
h = 31.752 m + 51.43776 m
h = 83.18976 m
Therefore, the maximum height of the puck above ice level is approximately 83.2 m above ice level, or 8.6 m above ice level if we consider the starting point as reference.
To solve this problem, we can break it down into steps:
Step 1: Calculate the vertical component of the puck's initial velocity.
Given:
- Angle of the shot, θ = 65°
- Time in the air, t = 1.8 s
The vertical motion of the puck can be described by the following equation:
Δy = v0y * t + (1/2) * g * t^2
Where:
Δy = change in vertical position (7.5 m)
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2)
Using the equation, we can rearrange it to solve for v0y:
v0y = (Δy - (1/2) * g * t^2) / t
v0y = (7.5 m - (1/2) * (-9.8 m/s^2) * (1.8 s)^2) / 1.8 s
v0y = (7.5 m + 16.74 m) / 1.8 s
v0y ≈ 13 m/s [up]
Therefore, the vertical component of the puck's initial velocity is approximately 13 m/s [up].
Step 2: Calculate the puck's initial velocity.
Using the vertical component of the initial velocity calculated in step 1 and the angle of the shot, we can calculate the initial velocity (v0) using the following trigonometric equation:
v0 = v0y / sin(θ)
v0 = 13 m/s / sin(65°)
v0 ≈ 14.29 m/s
Therefore, the puck's initial velocity is approximately 14.29 m/s, making an angle of 65° above the horizon.
Step 3: Determine the maximum height of the puck above ice level.
To determine the maximum height, we need to find the vertical displacement of the puck at the highest point of its trajectory.
Since the puck was shot upwards, the vertical component of its velocity at the highest point will be zero (v = 0). Using the equation:
v = v0y + g * t
0 = 13 m/s + (-9.8 m/s^2) * t
t = 13 m/s / (9.8 m/s^2)
t ≈ 1.33 s
Now, we can determine the maximum height using the formula:
Δy = v0y * t + (1/2) * g * t^2
Δy = 13 m/s * 1.33 s + (1/2) * (-9.8 m/s^2) * (1.33 s)^2
Δy ≈ 8.57 m
Therefore, the maximum height of the puck above ice level is approximately 8.57 m.
In summary:
(a) The vertical component of the puck's initial velocity is approximately 13 m/s [up].
(b) The puck's initial velocity is approximately 14.29 m/s, making an angle of 65° above the horizon.
(c) The maximum height of the puck above ice level is approximately 8.57 m.
To calculate the vertical component of the puck's initial velocity, we can use the equation of motion:
Vertical displacement (Δy) = initial vertical velocity (Vy0) * time (t) + (1/2) * acceleration (a) * time squared (t^2)
In this case, the vertical displacement is 7.5 m, the time is 1.8 s, and we know that the acceleration due to gravity is approximately 9.8 m/s^2.
Using the equation, we can rearrange it to solve for the initial vertical velocity:
Vy0 = (Δy - (1/2) * a * t^2) / t
Substituting the given values:
Vy0 = (7.5 m - (1/2) * 9.8 m/s^2 * (1.8 s)^2) / 1.8 s
Simplifying the equation:
Vy0 = (7.5 m - 15.876 m) / 1.8 s
Vy0 = -8.376 m / 1.8 s
Vy0 ≈ -4.65 m/s
Since the upward direction is positive, the vertical component of the initial velocity is 4.65 m/s up.
To calculate the puck's initial velocity, we can use the following formulas:
Vx0 = V0 * cos(θ)
Vy0 = V0 * sin(θ)
Where V0 is the initial velocity of the puck and θ is the launch angle of 65°.
Using these formulas, we can solve for V0:
V0 = √(Vx0^2 + Vy0^2)
V0 = √((V0 * cos(θ))^2 + (V0 * sin(θ))^2)
V0 = √(V0^2 * cos^2(θ) + V0^2 * sin^2(θ))
V0 = √(V0^2 * (cos^2(θ) + sin^2(θ)))
V0 = √(V0^2 * 1)
V0 = V0
Since there is no other component involved in V0 other than itself, we can conclude that the initial velocity is the same as the initial velocity of the puck. Therefore,
V0 = 4.65 m/s
Thus, the initial velocity of the puck is 4.65 m/s at an angle of 65° above the horizon.
To determine the maximum height of the puck above ice level, we can use the equation:
Vertical displacement (Δy) = initial vertical velocity (Vy0) * time (t) + (1/2) * acceleration (a) * time squared (t^2)
But in this case, the displacement is the maximum height, and we know that at the maximum height, the vertical velocity is zero, so Vy0 = 0.
Thus, the equation becomes:
Δy = (1/2) * a * t^2
Substituting the known values:
Δy = (1/2) * 9.8 m/s^2 * (1.8 s)^2
Δy = (1/2) * 9.8 m/s^2 * 3.24 s^2
Δy = 3.132 m^2/s^2
Taking the square root of the result, we find that the maximum height is approximately:
Δy ≈ 1.77 m
Therefore, the maximum height of the puck above the ice level is approximately 8.6 m.