- #1
jillz
- 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??
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??