Question

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

Answer 1

[AI-generated answer — Explain Bot]

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.