Projectile Motion: Stunt Penguin Flying High!
How long will the stunt penguin stay in the air?
If the stunt penguin is shot horizontally out of a cannon with an initial velocity of 20.0 m/s and the cannon is 5.00 m above the floor, how much time will the penguin spend in the air?
Answer:
In this physics problem, we can calculate the time the penguin spends in the air using the kinematic equation for vertical motion and the height from the ground.
The time for projectile motion is determined by the vertical motion. With an initial vertical velocity of 0 m/s and a height of 5.00 m above the floor, we can use the kinematic equation:
t = sqrt(2h/g)
Where:
- t = time in seconds
- h = height in meters (5.00 m)
- g = acceleration due to gravity (9.81 m/s²)
By plugging in the values, we can calculate the time the stunt penguin stays in the air before hitting the ground.
How far will the stunt penguin travel before hitting the ground?
Considering the initial velocity of 20.0 m/s, how far across the floor will the stunt penguin go before it lands?
Answer:
To determine the horizontal distance the penguin will travel, we can use the initial horizontal velocity and the time calculated in the previous part.
The horizontal distance (range) can be calculated using the equation:
x = uxt
Where:
- x = horizontal distance in meters
- ux = initial horizontal velocity (20.0 m/s)
- t = time in seconds (calculated previously)
By substituting the values, we can find out how far the stunt penguin will travel before hitting the ground. Get ready for an exciting ride with our adventurous penguin!