A clown is juggling cakes vertically up into the air. One of the cakes is thrown

Question

A clown is juggling cakes vertically up into the air. One of the cakes is thrown vertically upward from 1.40 m above the ground with an initial velocity of 6 m/s. He fails to catch the cake and it hits the ground.
Assume that acceleration due to gravity is 10m/s2 and no air resistance
A) How high above the ground is the cake at its highest point? (8p)
C) What is the velocity of the cake at its when it hits the ground? (8p)
D) How long is the cake in the air from the time it leaves the clowns hand to when it hits the ground? (10p)

Explanation / Answer

givens: y1 = 1.40 m vi = 6 m/s A) at the highest point, the final velocity of the cake will be zero vf^2 = vi^2 + 2a(d) 0^2 = 6^2 + 2(-10)(d) 20 d = 36 d = 1.8 m therefore, the highest point above the ground will be: d + y1 = 1.8 + 1.4 = 3.2 m C) overall, the cake is displaced 1.4 m downward from where it starts vf^2 = vi^2 + 2ad vf^2 = 6^2 + 2(-10)(-1.4) vf = 8 m/s (downward) D) vf = vi + at -8 = 6 + (-10)(t) 10t = 14 t = 1.4 seconds BOL

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