Question
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)
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)
Answers
GPT 3.5
(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.
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.