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Angle of inclination of the projectile: θ

  • Thread starter jillz
  • Start date
  • #1
15
0
d = at + ½ a t^2

Horizontally: 7.12m = a cos θ t

at = 7.12m/ cos θ

Vertically: 10m = a sin θ t + ½ g t^2

10m = (7.12m/ cos θ ) *sin θ + ½ g t^2 = 7.12m tan θ + ½ g t^2

10m = 7.12m tan θ + ½ g t^2

10m = 7.12m tan θ + ½ g 1.1^2g

Solve for θ

Angle of inclination of the projectile: θ = 29.8o

This answer (29.8) is wrong...that much I know; what did I do/where did I go wrong in my calcs??
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,056
72
Well, what's the question? What is "a"-- acceleration? Your first equation is incorrect, if so: d=ut+(1/2)at^2
 
  • #3
15
0
that's interesting; my teacher says to use a for acceleration... what is 'u'
 
  • #4
cristo
Staff Emeritus
Science Advisor
8,056
72
u is initial speed, sometimes denoted v_i. One does normally use a for acceleration, but that equation is not correct. See http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html [Broken] for the kinematic equations.
 
Last edited by a moderator:
  • #5
15
0
'u' is the variable for what??

if 'a' is acceleration and I'm not supposed to use 'u', then shouldn't the equation be d=at+(1/2)at^2 ??
 
  • #6
15
0
ok, thanks!
 

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