A hockey player lobs a puck at an empty net with an initial velocity of 8.00 m/s at an angle of 26.0°. What is the maximum height the puck reaches?

h = (v sinθ)^2 / 2g

Well, I'm not an expert in physics, but I am an expert in clowning around! Let me take a wild guess and say that the maximum height the puck reaches is somewhere around... Cloud 9! Because what better way to reach new heights than with a puck and a good ol' game of hockey, right?

To find the maximum height the puck reaches, we can use the principles of projectile motion.

The motion of the puck can be separated into horizontal and vertical components. The horizontal component of velocity remains constant throughout the flight, while the vertical component of velocity changes due to the effect of gravity.

To find the maximum height, we need to determine the time it takes for the puck to reach its peak in the vertical direction. We can use the equation:

v_yf = v_yi - g * t

Where:
v_yf is the final vertical component of velocity (which becomes zero at the peak)
v_yi is the initial vertical component of velocity
g is the acceleration due to gravity
t is the time of flight

In this case, the initial vertical component of velocity is given by:

v_yi = v * sin(theta)

Where:
v is the initial velocity of the puck
theta is the angle at which the puck was launched

Let's substitute the known values into the equation to find v_yi:

v_yi = 8.00 m/s * sin(26.0°)
= 8.00 m/s * 0.4384
= 3.5072 m/s

Next, we can use the equation to find the time of flight:

0 = 3.5072 m/s - 9.8 m/s^2 * t

Solving for t:

t = 3.5072 m/s / 9.8 m/s^2
= 0.3578 s

Therefore, it takes approximately 0.3578 seconds for the puck to reach its peak in the vertical direction.

Now we can calculate the maximum height using the vertical motion equation:

y = v_yi * t - (1/2) * g * t^2

Substituting the known values:

y = (3.5072 m/s) * (0.3578 s) - (1/2) * 9.8 m/s^2 * (0.3578 s)^2
= 1.252 m

Therefore, the maximum height the puck reaches is approximately 1.252 meters.