A projectile is lauched at some speed at an angle of 62 degrees and travels a ra

Question

A projectile is lauched at some speed at an angle of 62 degrees and travels a range of 244 meters. The launch angle is then changed, but the projectile is fired at the same speed as the original case and also travels the same range of 244 meters. Find the maximum vertical displacement at 62 degrees, (ymax1), and find the maximum vertical displacement for the second angle (ymax2), and then find the difference between them in meters (diff = ymax1 - ymax2).

Explanation / Answer

x=244m, a=62 deg, ay=-g x=v*t (vx doesnt have any accelaretion) 244=vx*t t=244/vx y=v0y*t-gt2/2>>> 0=v0y*244/vx-5*(244)2/vx2>>>> 0=v0y-5*244/vx 244/vx=0 (probably not) v0y-5*244/vx=0 v0y=5*244/vx now becouse of trio: vy=sin62*vx vx2*sin62=5*244>>>vx2=5*244/sin62 vx=1381.735m/s=4974.24km/h vy=sin62*vx=4392km/h v2=vx2+vy2>>>>>v=6635.71km/h

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