Acceleration due to Gravity/Terminal Speed Problem

In summary, a flowerpot is dropped from a balcony 28.5 m above the ground while a ball is thrown vertically downwards from a balcony 26 m above the ground one storey below. The initial velocity of the ball is 12.0 m/s [down]. The question is whether or not the ball passes the flowerpot before striking the ground and if so, how far above the ground are the two objects when the ball passes the flowerpot. Using appropriate equations for the motion of the objects and considering the different starting times and elevations, it is determined that the final velocity of the ball is 26 m/s and it hits the ground after 14 seconds. The delta y for the first second is 4.9
  • #1
CityNoise
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Hi, i am in a Grade 12 Physics class and i was hoping someone could help me out on a homework question:

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"A Flowerpot is dropped from a balcony of an apartment, 28.5 m above the ground. At a time of 1.00 seconds after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26 m above the ground. The initial velocity of the ball is 12.0 m/s [down]. Does the ball pass the flowerpot before striking the ground? if so, how far above the ground are the two objects when the ball passes the flowerpot?"
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any help would be greatly appreciated :smile:
 
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  • #2
We respectfully request that students make an attempt at solving the problem before asking for assistance.

Please try to write the appropriate equations for the motion of the flower pot, and the ball.

The flower pot is simply falling, whereas the ball is thrown with an initial velocity. One also must consider that ball is thrown 1 second later, so one must determined how to related the different times. And the flower pot and ball start at different elevations.
 
  • #3
Well i found out that the final velocity for the ball was 26 m/s and the total time for the ball to hit the ground would be 14 seconds using d=V1 + V2/2 * delta T for time. I also found that the delta y for the 1 second duration was 4.9 metres. So i found that the flowerpot is 23.6 metres from the ground when the ball is released at its starting position, 26 metres from the ground. I just don't know where to go from their.
 
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FAQ: Acceleration due to Gravity/Terminal Speed Problem

What is acceleration due to gravity?

Acceleration due to gravity is the acceleration that an object experiences due to the gravitational pull of the earth. It is a constant value of approximately 9.8 meters per second squared (m/s^2).

How is acceleration due to gravity calculated?

Acceleration due to gravity is calculated using the formula a = g = F/m, where a is acceleration, g is the acceleration due to gravity, F is the force of gravity, and m is the mass of the object.

What is terminal speed?

Terminal speed is the maximum speed that an object can reach when falling through a fluid, such as air. It occurs when the force of gravity equals the force of air resistance, resulting in a constant velocity.

How is terminal speed affected by an object's mass?

The mass of an object does not directly affect its terminal speed. However, a larger mass may require a greater force of air resistance to reach terminal speed. Objects with a larger surface area, such as a parachute, will have a lower terminal speed due to increased air resistance.

Why is understanding acceleration due to gravity and terminal speed important?

Understanding acceleration due to gravity and terminal speed is important in many fields, including physics, engineering, and aviation. It allows us to accurately predict the motion of objects and design structures and vehicles that can safely withstand gravitational forces. Additionally, understanding these concepts can help us understand the effects of gravity on our daily lives, such as the speed at which objects fall or the forces acting on our bodies during activities like skydiving or bungee jumping.

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