Projectile on horizontal surface

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Homework Help Overview

The problem involves a particle projected at a speed of 98 m/s at an angle A to the horizontal, with a specified range of 940.8 m. The goal is to find the two possible angles A.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations of motion in both horizontal and vertical directions, attempting to relate the range to the angle of projection. There are mentions of potential math errors in their calculations and attempts to derive the correct relationships between the variables.

Discussion Status

The discussion reveals that participants are actively identifying and correcting errors in their mathematical reasoning. Some guidance has been offered regarding the equations used, but no consensus has been reached on the correct approach or solution.

Contextual Notes

Participants express uncertainty about the derivation of equations and the implications of their calculations, indicating a need for clarification on the relationships between the variables involved.

Darth Frodo
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Homework Statement



A particle is projected with a speed of 98 m s−1 at an angle A to the horizontal.
The range of the particle is 940·8 m. Find
(i) the two values of A

The Attempt at a Solution



U = 98cosAi + 98sinAj

i direction

s = ut
940.8 = 98cosAt

j direction

s = ut + (0.5)at[itex]^{2}[/itex]
0= 98sinAt - (0.5)gt[itex]^{2}[/itex]
98sinA = gt/2
t = 20sinA

i direction
940.8 = (20sinA)(98cosA)
94.8 = (2sinAcosA)g
94.8 / g = sin2A



And I end up with a math error. Help please.
 
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Darth Frodo said:
940.8 = (20sinA)(98cosA)
94.8 = (2sinAcosA)g

Oops.
 
Ooops, sadly it doesn't solve the problem.

94.08 = 2sinAcosA(9.8)
94.08 / 9.8 = sin2A
9.6 = sin2A

math error
 
Darth Frodo said:
94.08 = 2sinAcosA(9.8)

This equation doesn't follow from 940.8 = (20)sinAcosA(98)
 
AHHH! Stupid error! Thanks!
 

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