Quoting Post Archived Topic: It will not be bumped to the top of the forum. Topic Physics Problem Membername   Not a member? Sign Up Free (takes 20 seconds) Password   Forgotten your password? Post Font: Verdana Arial Black Georgia Comic Sans Impact   Size: Tiny Normal Big Huge   Color: Default Dark Red Red Orange Brown Green Olive Cyan Blue Dark Blue Indigo Black [ Check Spelling ] [ Post Preview ] [ LiveTags ] Quote: from [m]morgan82[/m] at 4:49 pm on Jan. 4, 2009[quote]An object is launched with an initial velocity of 46 m/s at and angle of 57.0 degrees. What is the max height it will reach? Can someone help me out?[/quote] FAQ Keyword Search: Post Options Favorites Manager Notify me of new replies to this topic by email Notify me of new replies to this topic by private message Original Post morgan82 Posted at 4:49 pm on Jan. 4, 2009 An object is launched with an initial velocity of 46 m/s at and angle of 57.0 degrees. What is the max height it will reach? Can someone help me out?

Replies nullPointerException Posted at 5:13 pm on Jan. 4, 2009 y component of velocity = vsin(angle from horizontal)   Y compnentof velocity = (46m/s)(sin(57)) (V final)^2 = (V initial)^2 +2ad ((V final)^2 - (inital v of y component)^2)/2a = d ((0 m/s)^2 - (46m/s)(sin(57))^2) / (-2g) = d ((0 m/s)^2 - (46m/s)(sin(57))^2) / (-2 (9.81 m/s^2) )= d solve, dont have a calculator on hand DTXBrian Posted at 5:08 pm on Jan. 4, 2009 To be able to determine the maximum height of the object, you have to know how fast the initial vertical velocity is.  To do that, you take the sine of 57 degrees, which is .8387 and multiply it by the total initial velocity, and you get that the object is travelling approximately 38.6 m/s vertically. Now, you go back to your standard kinematic equations... The maximum height will be reached at the exact point the object is no longer traveling upwards, so using the equation V=Vo + AT, you can solve for T (time) 0=38.6 + -9.8T 9.8T = 38.6 9.8T/9.8 = 38.6/9.8 T = 3.94 seconds Now, go back to the equations... S = So + VoT + 1/2AT^2 S = 0 + 38.6x3.94 + .5x-9.8x3.94^2 S = 152.8 + -4.9x15.52 S = 152.8 - 76.05 S = 76.75 meters in the air. If you need more clarification, message me. blufindr Posted at 5:02 pm on Jan. 4, 2009 From the ground, right? sin(57) = y/46 y=46*sin(57) y=38.58ms^-1 You know the velocity at the top is going to be 0ms^-1 and acceleration is -9.8ms^-2 So v^2=u^2+2ax 0=148.83-19.6x x=148.83/19.6 x=75.94m So the object goes up 75.94 metres. Anonymous Posted at 4:57 pm on Jan. 4, 2009 hopefully this is right.   You gotta find the y component of the velocity, so you would do sin(57) x 46 m/s.  Then you get the y component speed.  Then you need to find the total kinetic energy the object has, using the equation 1/2(mass)(velocity)^2.  You don't have mass, but it will cancel out later anyway, so its ok. So then you have the kinetic energy which is   "blahblahblah(mass)"  and you set that equal to (mass)(accel due to gravity)(height).  The kinetic energy must equal the potential energy at the highest point. So the masses from each side of the equation cancel out. Then you get some number = (accel due to gravity)(height). So then you divide the left by the accel due to gravity, and you get the height sorry i don't have a calculator so i can't do the sin 57 part. bass09 Posted at 4:55 pm on Jan. 4, 2009 ya   about that commentary Posted at 4:53 pm on Jan. 4, 2009 I know it has to do with the sin of the 57 degree angle.. Fearfulteen Posted at 4:52 pm on Jan. 4, 2009 i'm only in 8th grade!!!!! snowfish Posted at 4:51 pm on Jan. 4, 2009 if you have a graphing calculator, you can just graph it with and find the peak of the parabola. Otherwise I think you break it down into vertical and horizontal components and solve it like a vector. I'm sorry I haven't taken a physics class in two years. Frubeling Posted at 4:51 pm on Jan. 4, 2009 SHIT, Ive got an exam on this in just over a week, I CANT DO IT YET All 9 previous replies displayed.