Calculating Time of Flight for a Launched Object

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To calculate the time of flight for an object launched horizontally from a height of 1.3 meters with an initial speed of 4.2 m/s, the correct approach involves using the equations for horizontal and vertical motion. The time of flight can be determined using the formula for vertical motion, d = vit + 1/2at^2, where the initial vertical velocity is zero and acceleration is due to gravity. A common method is to create an X and Y table to organize the known variables, ensuring that time remains constant across both dimensions. The correct time of flight for this scenario is 0.52 seconds, highlighting the importance of correctly applying the physics principles involved. Understanding these concepts can significantly improve problem-solving in physics.
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Homework Statement


A system is setup to launch an object horizontally off desks with an initial speed of 4.2 m/s. Height of desk is 1.3 meters.

What time of flight should be predicted for the object?

Homework Equations


Δt = (v2 - v1) / ∂ (Not too sure)

The Attempt at a Solution


1.3/4.2 = 0.31 seconds

Hello. I tried to find the time, but I keep getting the time wrong. The actual answer is 0.52s, but I think I'm going the wrong way about it :( Can someone help me understand the question more and lead me into the right direction? I'm not very good at physics.
 
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Are you aware of the X and Y table method?

Make an X and Y table
Like this

X___|___Y

In the X column, put all of your X axis information (you have velocity)

And in the Y column put down all of your Y axis information (you always have acceleration due to gravity, you have height/distance in this question as well)Now, remember that for horizontal you always use d = tv, and for vertical it's always d = vit + 1/2at^2.

Use those formulae with the information you have in the respective columns to find the unknown variables.

ALWAYS REMEMBER-
Time will be equal in both columns, it is the only factor that does not change because it's neither an X nor a Y nor a Z vector, so if you get time in one column you will effectively have time in both columns.
 
Last edited:
Ohh..! Thank you so much! You saved me!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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