Question #17781

A cannon sits atop a hill that is 10.5 m above the field below. The cannon shoots a shell with a velocity of 67 m/s at an angle of 54o. How far from the cannon will the shell hit the ground?

Expert's answer

Vertical velocity of a cannonis v*sin54o. Time, when it reaches the top of it's trajectory is

t=(v*sin54o)/g=5.53s. The height it's gonna reach is

H=H0+vtsin54o-gt^2/2=160.25m. Then it starting to fall freely with acceleration

of g. Time it's need to fall is t=sqrt(2H/g)=5.72s. So, the time it was in the

air is 5.53+5.72=11.25s. Horizontal velocity is v*cos54o. Distance it will cover

until it falls S=11.25*v*cos54o=260.41m.

t=(v*sin54o)/g=5.53s. The height it's gonna reach is

H=H0+vtsin54o-gt^2/2=160.25m. Then it starting to fall freely with acceleration

of g. Time it's need to fall is t=sqrt(2H/g)=5.72s. So, the time it was in the

air is 5.53+5.72=11.25s. Horizontal velocity is v*cos54o. Distance it will cover

until it falls S=11.25*v*cos54o=260.41m.

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