My Math Forum Velocity of a projectile

 April 22nd, 2016, 12:53 AM #1 Newbie   Joined: Apr 2016 From: Singapore Posts: 1 Thanks: 0 Velocity of a projectile Hi Guys, 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 Angle: 45deg a: -9.81m/s2 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): Total height = sqr [ Vr2 / 2(-9.81m/s2) ] = sqr [ (-930m/s)2 / ( -19.62m/s2) ] = sqr [ -864,900m2/s2 / -19.62m/s2 ] = sqr 44,082m = 210m 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 Senior Member   Joined: Apr 2014 From: UK Posts: 965 Thanks: 342 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)

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