Mapoob Sapling Learning this question was written by John Carlson at University
ID: 1870776 • Letter: M
Question
Mapoob Sapling Learning this question was written by John Carlson at University of Colorado, Denver A cannon, located 47.0m from the base of a vertical 26.6m tall cliff, shoots a 15.0kg shell at 40.8° above the horizontal toward the cliff Number What must the minimum muzzle velocity be for the shell to clear the cliff edge? m/s The ground at the top of the cliff is level, with a constant elevation of 26.6m above the cannon How far past the cliff edge does the shell land when the shell is fired at that minimum speed? Number Number What is the the shell just before it hits the ground? nitude of the acceleration of m/s For this launch angle, what is the frthest distance to the cliffs base that the cannon can be moved to and still reach the level ground at the cliff's top given this launch angle and velocity? Number ITmExplanation / Answer
time taken to reach the cliff,
47 = (v0 cos40.8 )(t) { in horizontal}
v0 t = 47 / cos40.8 ..... (i)
in vertical,
y = v0y t + ay t^2 / 2
26.6 = (vo sin40.8) (47 / cos40.8 v0) + (-9.8 t^2/ 2)
t = 1.69 sec
v0 = 36.8 m/s .............Ans
26.6 = (36.8 sin40..8) t - 9.8 t^2 /2
4.9 t^2 - 24 t + 26.6 = 0
t = 3.20 sec
R = (v0x) t = 36.8 cos40.8 x 3.20 = 89.1 m
past the edge = 89.1 - 47 = 42.1 m ...........Ans
a = 9.8 m/s^2 .........Ans
distance can be moved = 42.1 m .....Ans