- #1

- 15

- 0

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??