A cannonball is hot straight up from the edge of a cliff with an initial velocity of 30m/s. A second cannonball is shot straight downward from the edge of the same cliff with an initial velocity of -30m/s. How do the speeds of the two cannonballs compare when they reach the ground?
they are the same. same initial energy, same final energy
To compare the speeds of the two cannonballs when they reach the ground, we need to consider their motion.
The first cannonball is shot straight up with an initial velocity of 30 m/s. We can use the laws of motion to determine its final velocity when it reaches the ground. In this case, the acceleration due to gravity will act in the opposite direction of the cannonball's initial velocity, causing it to slow down.
The second cannonball is shot straight downward with an initial velocity of -30 m/s. Again, using the laws of motion, we can determine its final velocity when it reaches the ground. In this case, the acceleration due to gravity will act in the same direction as the cannonball's initial velocity, accelerating it further.
So, to compare the speeds of the two cannonballs, we can assume that the cliff is not too high, so the effects of air resistance can be neglected. Therefore, both cannonballs will experience the same acceleration due to gravity, which is approximately 9.8 m/s².
For the first cannonball:
Initial velocity (upward) = +30 m/s
Acceleration = -9.8 m/s² (opposite direction)
Using the following equation of motion:
v² = u² + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement (in this case, the height at which the cannonball reaches the ground)
We'll assume that "s" is positive in the upward direction, so it will be negative when the cannonball reaches the ground.
For the first cannonball:
u = 30 m/s (upward)
a = -9.8 m/s²
s = unknown (we'll solve for it)
v² = u² + 2as
v² = (30 m/s)² + 2(-9.8 m/s²)(-s)
Since "s" is negative, we can remove the negative signs:
v² = 900 + 19.6s
Now, let's consider the second cannonball:
Initial velocity (downward) = -30 m/s
Acceleration = +9.8 m/s² (same direction)
For the second cannonball:
u = -30 m/s (downward)
a = 9.8 m/s²
s = unknown (we'll solve for it)
v² = u² + 2as
v² = (-30 m/s)² + 2(9.8 m/s²)(-s)
Again, since "s" is negative, we can remove the negative signs:
v² = 900 + 19.6s
Comparing the equations for the first and second cannonballs, we can see that their final velocities will be the same (v²) if the values of "s" are equal. This means that when both cannonballs reach the ground, their speeds will be the same, regardless of their initial velocities.
In conclusion, when both cannonballs reach the ground, their speeds will be equal.