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.
horrizontal distance=vi*cosTheta*time
vi*costheta=12/.650
now, for time in the air
hf=hi+1/2 g t^2+ vi*sinTheta*t
2.8= -4.9 t^2+vi*sinTheta*t
you know t, for vi put in 12/(cosTheta*.65)
2.8= -4.9 t^2+ sinTheta*t*12/(cosTheta*.65)
now you have an equation you can solve for Tan Theta, then theta, the launch angle.
Then, you can go back and solve for vi, and hf max (when vvertical is zero).
To find the initial speed of the puck, we can use the horizontal distance and the flight time.
a.) The horizontal velocity of the puck is constant, so we can use the equation:
horizontal velocity = horizontal distance / time
horizontal velocity = 12.0 m / 0.650 s
horizontal velocity ≈ 18.46 m/s
b.) To find the maximum height the puck will reach, we can use the vertical motion equations.
First, we need to find the initial vertical velocity. Since the puck just clears the top of the wall, it means the vertical displacement is equal to the height of the wall (2.80 m) and the time taken to reach this point is half of the total flight time (0.650 s / 2 = 0.325 s).
Using the equation:
vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
2.80 m = initial vertical velocity * 0.325 s + (1/2) * (-9.8 m/s^2) * (0.325 s)^2
Simplifying the equation:
2.80 m = 0.325 * initial vertical velocity - 0.168 * 9.8 m
2.80 m = 0.325 * initial vertical velocity - 1.646
0.325 * initial vertical velocity = 4.446
initial vertical velocity ≈ 13.674 m/s
Now, we can find the maximum height the puck will reach using the equation:
maximum height = (initial vertical velocity)^2 / (2 * acceleration)
maximum height = (13.674 m/s)^2 / (2 * (-9.8 m/s^2))
maximum height ≈ 9.03 m
Therefore:
a.) The initial speed of the puck is approximately 18.46 m/s.
b.) The maximum height it will reach is approximately 9.03 m.
To find the initial speed of the puck, we can use the horizontal distance and flight time.
a.) The initial speed of the puck can be calculated using the horizontal distance and flight time with the formula:
Initial speed = horizontal distance / flight time
Substituting the given values:
Initial speed = 12.0 m / 0.650 s
Initial speed = 18.46 m/s (rounded to two decimal places)
Therefore, the initial speed of the puck is approximately 18.46 m/s.
To find the maximum height the puck will reach, we need to use the projectile motion equations. In this case, we are only considering the vertical motion of the puck.
We can use the equation:
Maximum height = (final vertical velocity)^2 / (2 * gravitational acceleration)
However, to find the final vertical velocity, we need to find the time it takes for the puck to reach its highest point.
The time to reach the highest point is half of the total flight time.
t_highest = flight time / 2
t_highest = 0.650 s / 2
t_highest = 0.325 s
Now we can find the final vertical velocity at the highest point using the equation:
Final vertical velocity = initial vertical velocity + (gravitational acceleration * time)
Since the puck is launched horizontally, there is no initial vertical velocity.
Final vertical velocity = gravitational acceleration * time to highest point
Final vertical velocity = 9.81 m/s^2 * 0.325 s
Final vertical velocity = 3.18825 m/s (rounded to five decimal places)
Now we can substitute this value into the equation to find the maximum height:
Maximum height = (3.18825 m/s)^2 / (2 * 9.81 m/s^2)
Maximum height = 0.5131 m (rounded to four decimal places)
Therefore, the maximum height the puck will reach is approximately 0.5131 m.