a rifle is fired from the top of a building 220 meters high with an initial velo

Question

a rifle is fired from the top of a building 220 meters high with an initial velocity of 425 m/s at an angle 12.0 degrees above the horizontal.

a. how long will it take for the bullet to reach the ground?

b. how far from the base of the building will the bullet strike the ground?

c. what will be the velocity of the bullet just as it reaches the ground?

i figured out the horizontal and vertical components of the initial velocity but have no idea where to continue from there

Explanation / Answer

vo=425 m/s--->voy=vo*sin(12 degrees)=88.36 m/s--->vox=v0*cos(12 degrees)=415.71 m/s max ht t:voy/g=9.02 s--->max ht:voy^2/(2*g)=220+88.36^2/(2*9.8)=618.4 m ((-voy+sqrt(88.36^2+4*1/2*9.8*220)^(1/2))/(-2*4.9)=20.25 s+9.02*2=38.28 s total flight time:38.28 s a) 38.28 s for bullet to hit ground level b)vox*t=15914.7 m from base of building c) vy final:88.36+9.8*38.28=463.54 m/s--->vox=vx final=415.71 m/s sqrt(463.54^2+415.71^2)=622.6 m/s--->final angle on ground:arctan(463.54/415.71)=48.11 degrees

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