Question

#3) A walnut is shot nearly vertically upward from a point near the edge of a tall building from a cannon. It just misses the edge on the way down (when it is in free fall) and passes a point a distance 40.0 m below its starting point at a time 5.00 s after it leaves the cannon barrel. Air resistance may be ignored. Report answers to the below questions to 3 sig figs. Set downward as the negative direction. A) What is the initial velocity of the walnut? B) How high does it rise above its starting point? C) What is the magnitude of its velocity at the highest point? D) What is the magnitude of its acceleration at the highest point? E) What is the direction of its acceleration at the highest point?

Answer 1

a)

using 2nd equation of motion

h = u t - 0.5 gt^2

40 = u * 5 - 4.9* 5^2

u = 32.5 m/s

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b)

H = u^2/2g = 32.5^2 / 19.6 = 53.89 m

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c)

v = 0 m/s

d)

a = 9.8 m/s^2

e)

in downward direction