Projectile on horizontal surface

AI Thread Summary
A particle is projected at a speed of 98 m/s at an angle A, with a range of 940.8 m, prompting a need to find two values of A. The calculations involve breaking down the motion into horizontal and vertical components, using equations of motion. An error occurs in the transition from the range equation to the sine double angle formula, leading to confusion in solving for A. The correct relationship should be established between the range and the trigonometric identities. The discussion highlights the importance of careful algebraic manipulation in projectile motion problems.
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^{2}
0= 98sinAt - (0.5)gt^{2}
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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