During a Science day learners use a plastic cold drink bottle as a rocket.They use a bicycle pump to increas the presure inside the bottle.When the presure becomes high enough,the bottle is launched vertically upwards into the air.The bottle returns to its original position after 3 second.
1.What is the magnitude of the velocity of the bottle when it reaches its maximum height?
2.What is the magnitude and the direction of the acceleration of the bottle during its flight?
1. 0 m.s-1
2. 9.8 m.s-2 downwards
To determine the magnitude of the velocity of the bottle when it reaches its maximum height, we can use the equation of motion for vertical motion.
1. First, we need to identify the known variables:
- Initial velocity (u) = 0 m/s (since the bottle was launched vertically upwards from rest)
- Acceleration (a) = -9.8 m/s^2 (assuming we're neglecting air resistance and taking the acceleration due to gravity)
- Time (t) = 3 sec (as given in the question)
2. Using the equation for vertical motion, which is:
v = u + a*t
where:
- v is the final velocity (magnitude of velocity) at maximum height
3. Substituting the known values into the equation:
v = 0 + (-9.8 m/s^2) * 3 sec
v = -29.4 m/s
Therefore, the magnitude of the velocity of the bottle when it reaches its maximum height is 29.4 m/s. Note that the negative sign indicates that the velocity is directed downwards.
Now, let's move on to the second question regarding the magnitude and direction of the acceleration of the bottle during its flight.
Since the bottle is flying vertically upwards and then returns to its original position, the acceleration experienced by the bottle is due to gravity and is constant at all points. The magnitude of the acceleration due to gravity is approximately 9.8 m/s^2 and always acts downwards.
Therefore, the magnitude of the acceleration of the bottle during its flight is 9.8 m/s^2, and its direction is downwards.