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
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.