A child is playing with a toy cannon that fires metal balls. The muzzle velocity

Question

A child is playing with a toy cannon that fires metal balls. The muzzle velocity of the cannon 400 cm/sec, and the cannon is set up to fire the cannon ball at an angle of 60 dehrees above the horizontal. Assume that the cannon ball is fired from ground level.

a) determine the maximum height above the ground that the cannon ball will reach in cm

b) If the cannon is fired over ground level ground with no obstructions, how far will the cannon ball travel before hitting the ground?

c) One afternoon, the child decides to fire the canon inside the house. He shoots the cannon toward the wall which is a distance of one meter away. The canon makes a hole in the wall when it strikes it. How high (in cm) is this hole above the floor?

Explanation / Answer

a)

Let initial angle of projection is p

Max. height = V^2*sin2(p)/2g

=> max. height = (4)^2*(sin(60))^2/2*9.8 = 0.61 m or 610 cm

b)

Range = V^2*sin2(p)/g = (4*4*0.865)/9.8 = 1.41m or 1410 cm

c)

equation of trajectory

Y = xtan(p) - gx^2/(2v^2*cos^2(p))

so let's calculate Y at x=1

=> Y = 1*tan(60) - 9.8*1*1/(2*4*4*0.5*0.5) = 1.73  - 9.8*1*1/(2*4*4*0.5*0.5)

=> Y = 0.505m or 505cm

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