A projectile of mass 0.757 kg is shot from a cannon, at height 6.7 m, as shown i

Question

A projectile of mass 0.757 kg is shot from a cannon, at height 6.7 m, as shown in the figure, with an initial velocity vi having a horizontal component of 6.7 m/s. The projectile rises to a maximum height of
y above the end of the cannon

Explanation / Answer

Given the angle of the trajectory and the horizontal component of muzzle velocity, we can determine the vertical component of muzzle velocity Vxo = 6.6ms = Vo*cos51 => 6.6/cos51 = Vo = 10.49m/s => Vyo=10.5sin51=8.15m/s a) The peak altitude occurs when Vy= Vyo -gt = 0 => t = Vyo/g = 0.83s How high did it go? Height = 6.7+Vyo*t-0.5gt^2 = 6.7+8.15*0.83-4.9*0.83^2=10.1m delta y = 10.1-6.7=3.39m b) How long does it take to fall from this height? 10.1 = gt^2 => t = sqrt(10.1/9.8) = 1.02s The total flight time is the time to peak altitude plus the time from peak altitude to ground 0.83+1.02= 1.85s How far did it go horizontally in 1.85s? 6.6*1.85=12.19m F) C The work done by gravity is weight mg times the movement inline with the force of the weight The work going up is therefore negative, the work going down is positive weight = 0.339*9.8=3.32N Distance up is 3.39m distance down is 10.1 3.32*(10.1-3.39)=22.3J C) What is the vertical velocity upon impact Vy=gt = 9.8*1.02 = 9.94ms The horizontal velocity is 6.6 The magnitude of the impact velocity is sqrt(6.6^2+9.94^2)=11.93m/s D) E THe angle on impact = arctan Vy/Vx =-9.94/6.6 =56.42 degrees with the negative x axis or 123.57 degrees with the positive x axis. E)

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