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Playing jackstone as a kid, you discovered that there is a certain time before the rubber ball falls and you took advantage of this when picking up the stars on the floor. If you threw the ball 4.10m/s up into the air, how fast will it be moving at a height of 42.45cm? How much time has elapsed upon reaching this height
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To answer your question, we can use the laws of motion, specifically the equations of motion for an object thrown vertically upwards. We'll use the following equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity of the ball at the given height (which we want to find)
- u is the initial velocity of the ball (which is 4.10 m/s upwards)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
- s is the height reached by the ball (which is 42.45 cm converted to meters, i.e., 0.4245 m)
Let's solve for v:
v^2 = u^2 + 2as
v^2 = (4.10 m/s)^2 + 2(-9.8 m/s^2)(0.4245 m)
v^2 = 16.81 m^2/s^2 + (-8.296 m^2/s^2)
v^2 = 8.514 m^2/s^2
Now, let's take the square root of both sides to find v:
v = √(8.514 m^2/s^2)
v ≈ 2.918 m/s
Therefore, the ball will be moving at approximately 2.918 m/s when it reaches a height of 42.45 cm.
To find the time elapsed upon reaching this height, we can use the equation:
v = u + at
Where:
- v is the final velocity (which is 2.918 m/s)
- u is the initial velocity (which is 4.10 m/s upwards)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
- t is the time elapsed
Now, let's solve for t:
v = u + at
2.918 m/s = 4.10 m/s + (-9.8 m/s^2) * t
2.918 m/s - 4.10 m/s = -9.8 m/s^2 * t
-1.182 m/s = -9.8 m/s^2 * t
t = (-1.182 m/s) / (-9.8 m/s^2)
t ≈ 0.121 seconds
Therefore, approximately 0.121 seconds will have elapsed upon reaching a height of 42.45 cm.