A hockey puck truck at ice level just clears the top of a glass wall that is 2.80 m high. The flight time to this point is 0.650 s, and the horizontal distance is 12.0 m. Find a.) the initial speed of the puck and b.) the maximum height it will reach.
To solve this problem, we can use the kinematic equations of motion. Let's start by finding the initial speed of the puck.
a.) Finding the initial speed (Vo):
We can use the horizontal distance (x) and the time of flight (t) to find the horizontal component of the initial velocity (Vox).
The formula for the horizontal distance is:
x = Vox * t
Rearranging the equation to solve for Vox:
Vox = x / t
Plugging in the given values:
Vox = 12.0 m / 0.650 s
Vox ≈ 18.46 m/s
So, the initial speed of the puck is approximately 18.46 m/s.
b.) Finding the maximum height (Hmax):
To find the maximum height, we can use the vertical component of the initial velocity (Voy) and the time of flight (t).
The equation for the maximum height can be derived from the kinematic equation for vertical displacement (y):
y = Voy * t + (1/2) * g * t^2
where g is the acceleration due to gravity.
At the maximum height, the final vertical velocity (Vy) is zero. So, we can use this information to write the equation:
0 = Voy + (g * t)
Rearranging this equation to solve for Voy:
Voy = -g * t
Plugging in the values:
Voy = -9.8 m/s^2 * 0.65 s
Voy ≈ -6.37 m/s
Here, the negative sign indicates that the vertical velocity is in the opposite direction to the positive y-axis (upward).
Now, using the formula for maximum height, we can find Hmax:
Hmax = (Voy^2) / (2 * g)
Plugging in the values:
Hmax = (-6.37 m/s)^2 / (2 * 9.8 m/s^2)
Hmax ≈ 2.05 m
So, the maximum height the puck will reach is approximately 2.05 m.
To solve this problem, we can use kinematic equations of motion. Let's break down the problem into two parts: the vertical motion and the horizontal motion.
a.) To find the initial speed of the puck, we'll focus on the horizontal motion. We can use the equation:
Horizontal distance (d) = Initial velocity (Vx) × Time (t)
Given that the horizontal distance is 12.0 m and the time is 0.650 s, we can rearrange the equation to solve for the initial velocity:
Vx = d / t
Vx = 12.0 m / 0.650 s
Vx ≈ 18.46 m/s
So, the initial speed of the puck is approximately 18.46 m/s.
b.) Now, let's find the maximum height reached by the puck. We'll focus on the vertical motion. The initial vertical velocity (Vy) is zero, as the puck is launched horizontally. The acceleration due to gravity (g) is approximately 9.8 m/s².
Using the kinematic equation for vertical motion:
Vertical displacement (d) = (Initial velocity (Vy) × Time (t)) + (1/2) × (Acceleration (g) × Time (t))²
At maximum height, the vertical displacement is equal to the height of the glass wall, which is 2.8 m. Rearranging the equation, we have:
2.8 m = (0 × t) + (1/2) × (9.8 m/s²) × t²
Simplifying:
2.8 m = 4.9 m/s² × t²
t² = 2.8 m / (4.9 m/s²)
t² ≈ 0.5714
t ≈ √0.5714
t ≈ 0.756 s
So, the time taken to reach maximum height is approximately 0.756 s. Now, we can find the initial vertical velocity (Vy) using another kinematic equation:
Vertical displacement (d) = (Initial velocity (Vy) × Time (t)) + (1/2) × (Acceleration (g) × Time (t))²
Rearranging the equation:
2.8 m = Vy × 0.756 s + (1/2) × (9.8 m/s²) × (0.756 s)²
Simplifying:
2.8 m = 0.756 s × Vy + 2.1816 m
0.756 s × Vy = -2.1816 m
Vy = (-2.1816 m) / (0.756 s)
Vy ≈ -2.89 m/s
By taking the negative sign, it indicates the initial vertical velocity is directed downward. Therefore, the maximum height reached by the puck is 2.89 m.
To summarize:
a.) The initial speed of the puck is approximately 18.46 m/s.
b.) The maximum height reached by the puck is approximately 2.89 m.