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Rjune1 Rjune1
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11 years ago
1) A cannonball is shot at 105 m/s at an angle of 20 degrees to the horizontal. Find the range (delta-x) and the max height the ball goes. Assume, for this problem, that the ball lands at the same vertical height that it is shot from.
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11 years ago
Part A) Max Height:

Initial vertical component of speed = (105 x sin 20) = 35.91 m/s

Rearrange and use SUVAT equation v^2 = u^2 + (2 x a x s),

s = (v^2 - u^2)/(2 x a), s = (0 - 1289.53)/(-19.62) = 65.723 metres

Part B) Range of the Cannon Ball

Rearrange and use SUVAT equation v = u + (a x t) to find time taken for ball to reach maximum height.  Then double it to find total time cannon ball was in the air (up and down times are equal)

t = (v - u) / a , t = (0 - 35.91)/ -9.81 = 3.661, therefore total t = 7.321 seconds

find the horizontal component of the initial speed and multiply it by the total time to find the range

(cos 20 x 105) x t = range, 98.668 x 7.321 = 722.35 metres
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