Angle of inclination of the projectile: θ

1. Mar 28, 2008

jillz

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??

2. Mar 28, 2008

cristo

Staff Emeritus
Well, what's the question? What is "a"-- acceleration? Your first equation is incorrect, if so: d=ut+(1/2)at^2

3. Mar 28, 2008

jillz

that's interesting; my teacher says to use a for acceleration... what is 'u'

4. Mar 28, 2008

cristo

Staff Emeritus
u is initial speed, sometimes denoted v_i. One does normally use a for acceleration, but that equation is not correct. See here for the kinematic equations.

5. Mar 28, 2008

jillz

'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. Mar 28, 2008

ok, thanks!