morbalt
In Angry Birds, your bird launching catapult is at the top of a tower 10m high. You launch your bird with a velocity of 10m/s at an angle of 30o. How far from the base of your tower will the bird land? Assume acceleration due to gravity is 9.81m/s2. Assume no friction
Help me please

AceDread
11/16/2012 7:23:47 AM Permalink[quote][b]Posted by:[/b] theHurtfulTurkey Vx = 10cos(30) You need to find t, the time it takes to hit the ground. You do this by finding Vy (10sin(30)), and using a = 9.81, and an additional height of 10m, solve for t. Using x = .5a(t^2) + v_0(t) + x_0 x_vertical = .5(9.81)(t^2) + (10sin(30))t + 10 x_horizontal = .5(0)(t^2) + (10cos(30))t + 0 Do you understand why acceleration for the horizontal distance is zero?[/quote] Damn son, you're smart. 
morbalt
11/16/2012 7:17:06 AM Permalink[quote][b]Posted by:[/b] theHurtfulTurkey Vx = 10cos(30) You need to find t, the time it takes to hit the ground. You do this by finding Vy (10sin(30)), and using a = 9.81, and an additional height of 10m, solve for t. Using x = .5a(t^2) + v_0(t) + x_0 x_vertical = .5(9.81)(t^2) + (10sin(30))t + 10 x_horizontal = .5(0)(t^2) + (10cos(30))t + 0 Do you understand why acceleration for the horizontal distance is zero?[/quote] thanks man figured it out 
HurtfulTurkey
11/16/2012 7:10:32 AM PermalinkVx = 10cos(30) You need to find t, the time it takes to hit the ground. You do this by finding Vy (10sin(30)), and using a = 9.81, and an additional height of 10m, solve for t. Using x = .5a(t^2) + v_0(t) + x_0 x_vertical = .5(9.81)(t^2) + (10sin(30))t + 10 = 0 x_horizontal = .5(0)(t^2) + (10cos(30))t + 0 Do you understand why acceleration for the horizontal distance is zero? [Edited on 11.15.2012 11:38 PM PST] 
morbalt
11/16/2012 7:08:44 AM Permalink[quote][b]Posted by:[/b] Sonkaro I just learned this in physics and already forgot haha, I would assume less then 30m in one second an object fall 5m in frictionless environment. d=1/2*g*t^2[/quote] I just can't remember the forumla or anything for it, Can't find the formula when a google it either 
