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how much does gravity affects my cannon shot?
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It depends on a lot of factors.
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A daredevil is shot out of a cannon at 49.7◦ to the horizontal with an initial speed of 29.9 m/s. A net is positioned at a
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hf=29.9*sinTheta*time-1/2 g t^2 distance: 49.8=29.9cosTheta*time solve for time, put it in the first
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A daredevil is shot out of a cannon at 44.9◦ to the horizontal with an initial speed of 27.5 m/s. A net is positioned at a
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H(x)=vi*t - 1/2 9.8 t^2 x=41.2 vi=27.5*sin44.9 now for t: d(x)=27.5*cos44.9*t but d(x)=41.2 solve
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i was wondering if i did this right
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The cannon on a pirate ships shoots cannon balls with a speed of 350m/s (the muzzle velocity). The cannon can be adjusted to
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Since the cannon ball is being shot horizontally, there is no vertical motion and therefore no
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The cannon on a pirate ships shoots cannon balls with a speed of 350m/s (the muzzle velocity). The cannon can be adjusted to
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The impulse delivered to the ship can be determined using the principle of conservation of momentum.
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The cannon on a pirate ships shoots cannon balls with a speed of 350m/s (the muzzle velocity). The cannon can be adjusted to
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vfinal at the cannon end is vo for the projectile motion.
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A cannon was shot off a cliff and traveled a distance of 498 m. Suppose the cannon was shot at a horizontal speed of 56 m/s.
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A large metal ball is shot from a cannon with a short barrel. If the same ball were to be shot from cannon with a longer barrel,
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d. is the only possible answer. Consider this, what if the barrel were extended two miles long,
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a shell is fired from the cannon. the shell of mass 30kg is shot with an initial velocity of 30m/s. what is the momentum of the
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To find the momentum of the cannon just after the shell is shot, we need to consider the principle
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