Simple Laws of motion question

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of an object launched vertically. The original poster describes an experiment where an object weighing approximately 550 grams was airborne for about 4.5 seconds, seeking to calculate its launch speed and peak height.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the effects of gravity on the object's motion, with one suggesting the use of kinematic equations. Others express confusion about the calculations and seek clarification on the relationships between time, velocity, and height.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the physics involved. Some guidance has been offered regarding the use of gravitational acceleration and the relationship between time and velocity, but there is no explicit consensus on the calculations presented.

Contextual Notes

Participants note the assumption of negligible air resistance and express uncertainty about the accuracy of their calculations due to a lack of prior physics experience.

faamfwvame
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A friend and I were conducting a science experiment, where an object weighing roughly 550 grams was launched vertically, and after examining the video footage, was airborne for aproximatly 4.5 secconds. Having never studied physics before, I was curious if you could calculate the launch speed and or peak height from this information.

Any help would be highly appreciated.
 
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Assuming negligible air resistance, which is approximatelytrue if the object is a ball made of a dense material, the velocity of a body moving against gravity is v = v_0 - gt
At the top of the trajectory the velocity is 0 and the time is half the total time.
 
I don't really understand that sorry, but according to Wikipedia, a stationary object's speed increases by 9.81 metres per second as it falls. and since my object was falling for 4.5/2, 2.25 secconds, I figured the peak hieght would be 9.81 + (9.81x2) + (9.81x3)/4 which equals 36.7875 metres, which, having seen the object, sounds about right. And 9.81 being approximatly 22 miles per hour, I guessed the speed it was traveling when it hit the ground and indeed the speed it was launched at would be 22 x 2.25 which is 49.5 miles per hour.
However, having no expirience with physics, I am hesitant to trust these results. Could anyone confirm or correct this?
Thank you for your input.
 
faamfwvame said:
I don't really understand that sorry, but according to Wikipedia, a stationary object's speed increases by 9.81 metres per second as it falls. and since my object was falling for 4.5/2, 2.25 secconds, I figured the peak hieght would be 9.81 + (9.81x2) + (9.81x3)/4 which equals 36.7875 metres, which, having seen the object, sounds about right. And 9.81 being approximatly 22 miles per hour, I guessed the speed it was traveling when it hit the ground and indeed the speed it was launched at would be 22 x 2.25 which is 49.5 miles per hour.
However, having no expirience with physics, I am hesitant to trust these results. Could anyone confirm or correct this?
Thank you for your input.
I don't understand the calculations you made. The initial and final velocities are equal in absolute values.
v_0 = 9.81*2.25 = 22m/s = 11.4 mph
The maximum heogth will be
h = v_0*t - \frac{1}{2]g*t^2 = 24.8m
 

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