|April 22nd, 2016, 12:53 AM||#1|
Joined: Apr 2016
Velocity of a projectile
First time post, I hope this is the correct location to post my 2 questions?
1/ I am trying to work out the horizontal distance a projectile will travel on level ground:
Resultant Velocity: 930m/s
I determined that Vx & Vy were 657m/s each.
I determined that tUp 67s, therefore tTotal was 134s
To calc the horizontal distance it should be just a matter of 134s x 657m/s (tTotal x Vx). Correct?
That gives me = 88,119m or 88km... this seems way off?
2/ Same projectile accelerates vertically (at 90deg from horizontal):
= sqr [ Vr2 / 2(-9.81m/s2) ]
= sqr [ (-930m/s)2 / ( -19.62m/s2) ]
= sqr [ -864,900m2/s2 / -19.62m/s2 ]
= sqr 44,082m
Dist = 420m (seems too little to me??)
tUp = 930m/s / -8.91m/s2 = 94.8s
tTotal = 189.6s (3min 3s)
Again it only travels 420m vertically but takes 3min & 3s to travel...?
I cannot see any errors but the values don't look like they line up. I have been staring at this for 20m.
Any help would be really appreciated before I tear all my hair out. Thanks.
Last edited by skipjack; April 22nd, 2016 at 02:42 AM.
|April 22nd, 2016, 02:37 AM||#2|
Joined: Apr 2014
1) Looks correct
2) After 1 second, the speed has dropped from 930m/s to 920.19m/s, so it must have travelled at least 920m in the first second alone, so yes, I think 420m is incorrect.
I see 44082m in your calculations, I get this number if I do 94.8 x 930/2 (flight time X average speed)
|Thread||Thread Starter||Forum||Replies||Last Post|
|projectile motion, how to get the equation for range in projectile motion||sweer6||New Users||4||May 21st, 2014 10:00 PM|
|projectile motion||shalini maniarasan||Calculus||1||April 25th, 2014 01:26 AM|
|Obtaining angular velocity given linear velocity and center of rotation 3D||quarkz||Calculus||0||April 18th, 2014 05:34 AM|
|Angle of projectile||Canning_s||Algebra||1||November 13th, 2010 12:43 PM|
|Projectile Motion||symmetry||Algebra||1||June 19th, 2007 10:26 PM|