Finding launch angle for projectile reaching 10m vertical and 7.12m horizontal

[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??

 

[cristo]

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

 

[jillz]

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

 

[cristo]

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 for the kinematic equations.

 

[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 ??

 

[jillz]

ok, thanks!