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Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 1.01 s after passing his window. Roger's room is on the third floor, 15 m
above the sidewalk.
Part A
Part complete
How fast are the balloons traveling when they pass Roger's window?
Express your answer to three significant figures and include the appropriate units.
Part B
Assuming the balloons are being released from rest, from what floor are they being released? Each floor of the dorm is 5.0 m high.
Express your answer as an integer.

Thus the balloons are coming from the floor number

Roger sees water balloons fall past his window He notices that each balloon strikes the sidewalk 101 s after passing his window Rogers room is on the third floo class=
Roger sees water balloons fall past his window He notices that each balloon strikes the sidewalk 101 s after passing his window Rogers room is on the third floo class=

Respuesta :

oladayoademosu

Part A The balloon is travelling at 19.8 m/s when it passes Roger's window

Part B The balloon was released from the 7th floor.

To solve the questions, we need to use the equations of motion

What are equations of motion?

Equations of motion are equations used to solve for the motion of an object under a constant acceleration

Part A How fast are the balloons traveling when they pass Roger's window?

Since Roger sees water balloons fall past his window and he notices that each balloon strikes the sidewalk 1.01 s after passing his window. Roger's room is on the third floor, 15 m above the sidewalk.

Using the equation of motion s = ut - 1/2gt² where

  • s = height of Roger's room above sidewalk = 15 m,
  • u = speed of balloon at Roger's window,
  • t = time it takes balloon to reach sidewalk = 1.01 s and
  • g = acceleration due to gravity = 9.8 m/s²

Making u subject of the formula, we have

u = s/t + 1/2gt

Substiting the values of the variables into th equation, we have

u = s/t + 1/2gt

u = 15 m/1.01 s + 1/2 × 9.8 m/s² × 1.01 s

u = 14.851 m/s + 1/2 × 9.898 m/s

u = 14.851 m/s + 4.949 m/s

u = 19.8 m/s

So, the balloon is travelling at a speed of 19.8 m/s when it passes Roger's window

Part B From what floor are they being released

Since the balloons are released from rest, they have zero velocity.

So, using the equation of motion

v² = u² - 2gh where

  • u = initial velocity of balloon = 0 m/s (since it is released from rest),
  • v = speed of balloon at Roger's window = 19.8 m/s,
  • g = acceleration due to gravity = 9.8 m/s² and
  • h = height which balloon was released

Making h subject of the formula, we have

h = (v² - u²)/2g

Substituting the values of the variables into ther equation,we have

h = (v² - u²)/2g

h = [(19.8 m/s)² - (0 m/s²)]/2(9.8 m/s²)

h = 392.04 m²/s²/19.6 m/s²)

h = 20 m

Since each floor is 5 m high, so a height of 20 m would be 20 m/5 m = 4 floors.

So, this is 4 floors above Roger's room.

Since roger's room is on the 3rd floor, then 4 floors + 3 floors = 7 floors.

So, the balloon was released from the 7th floor.

Learn more about equations of motion here:

https://brainly.com/question/27899272

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