Snowballs are thrown with a speed of 12.5 m/s from a roof 7.8 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 16° above the horizontal.
(a) Is the landing speed of snowball A greater than, less than, or the same as the landing speed of snowball B?
same speed greater than less than
Explain
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(b) Verify your answer to part (a) by calculating the landing speed of both snowballs.
snowball A m/s
snowball B m/s
help please. I don't need answers just a way to solve it
To solve this problem, we can break it down into two parts: determining the landing speed for each snowball and comparing the speeds.
(a) To determine whether the landing speed of snowball A is greater than, less than, or the same as snowball B, we need to understand the physics behind projectile motion.
When a projectile is launched at an angle above the horizontal, its initial velocity can be broken down into two components: horizontal and vertical. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
For snowball A, since it is thrown straight downward, its initial velocity is purely vertical. So, its horizontal component is zero.
For snowball B, it is thrown in a direction 16° above the horizontal. This means it has both horizontal and vertical components. The vertical component is affected by gravity, pulling the snowball downwards, while the horizontal component remains constant.
Now, let's think about the landing speed. The landing speed is the magnitude of the velocity vector just before impact, which includes both the horizontal and vertical components. Since snowball B has both horizontal and vertical components, its landing speed will be greater than that of snowball A, which only has a vertical component.
Therefore, the landing speed of snowball A is less than the landing speed of snowball B.
(b) To calculate the landing speed of each snowball, we can use the formula for projectiles:
Vertical component of velocity (Vy) = gt
Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.
For snowball A:
Since it is thrown straight downward, its initial velocity is in the negative y-direction (downwards). So, the vertical component of velocity for snowball A is -12.5 m/s (negative because it is downwards).
To calculate the time it takes for snowball A to reach the ground, we can use the formula:
Δy = Vy*t + (1/2)*g*t²
Where Δy is the vertical displacement and is equal to the height of the roof (7.8 m) since snowball A is dropped from the roof.
Rearranging the formula, we have:
7.8 m = -12.5 m/s * t + (1/2)*9.8 m/s² * t²
This is a quadratic equation that we can solve to find t. Once we have t, we can calculate the landing speed for snowball A by multiplying the time by the acceleration due to gravity:
Landing speed for snowball A = |Vy| = |(-12.5 m/s) + (9.8 m/s² * t)|
For snowball B:
Since it is thrown at an angle above the horizontal, we need to find the horizontal and vertical components of its initial velocity.
Horizontal component of velocity (Vx) = V * cos(θ)
Vertical component of velocity (Vy) = V * sin(θ)
Where V is the initial speed (12.5 m/s) and θ is the launch angle (16°).
To calculate the time it takes for snowball B to reach the ground, we can use the vertical component of velocity and the free fall equation:
Δy = Vy*t + (1/2)*g*t²
Where Δy is once again the vertical displacement and is equal to the height of the roof (7.8 m). Rearranging the formula, we have:
7.8 m = (V * sin(θ)) * t + (1/2)*g*t²
This is also a quadratic equation that we can solve to find t. Once we have t, we can calculate the landing speed for snowball B using the Pythagorean theorem:
Landing speed for snowball B = sqrt((V * cos(θ))² + (V * sin(θ) - g * t)²)
Using these formulas and solving the quadratic equations, you can find the landing speeds for both snowball A and snowball B.