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Originally posted by lauseyIf you can answer the following question you are ready for Walking 20...
Correct me if I am wrong, but doesn't one have to take Crawling 10 in playschool before Walking 20? The reason I am asking is because I recieved my schedule, and it has walking 20, when I haven't even taken the 10 yet.
A man throws a ball 10 feet into the air, while simultaneously a man 15 feet up drops a ball.
Initial velocity of the ball for man 1 = 3 m/s
Initial velocity of the ball for man 2 = 5 m/s
When are they both at the same height. What is the height that they meet?
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Originally posted by cmsMasterExcuse the notation considering subscripts and superscripts cannot be typed here.
If you can answer the following question you are ready for Walking 20...
A man throws a ball 10 feet into the air, while simultaneously a man 15 feet up drops a ball.
Initial velocity of the ball for man 1 = 3 m/s
Initial velocity of the ball for man 2 = 5 m/s
When are they both at the same height. What is the height that they meet?
Assuming man 1 throws the ball up from ground level (considering we do not know the height of the man):
v01 is initual velocity of ball 1
v1 is final velocity of ball 1
x01 is initual displacement of ball 1
x1 is final displacement of ball 1
v02 is initual velocity of ball 2
v2 is final velocity of ball 2
x02 is initual displacement of ball 2
x2 is final displacement of ball 2
a is accelaration
t is time
v01 = 3 ms^-1
v1 = 0
x0 = 0
v02 = -5 ms^-1
x02 = 15 feet = 15 x 12 x 2.54 / 100 = 0.709 m
a = -9.81 ms^-2
v01t + (1/2)at^2 + x01 = v02t + (1/2)at^2 + x02
3t - (1/2)*9.81t^2 = -5t - (1/2)9.81t^2 + 0.709
3t - 4.905t^2 = -5t - 4.905t^2 + 0.709
8t = 0.709
therefore, time for balls to meet:
t = 0.0886 seconds
x1 = 3 * 0.0886 + (1/2) * -9.81 * 0.0886
x1 = 0.266 - 0.435 = -0.169 m
hence 17 cm below the surface, maybe man 1 was standing over a man hole. 😕
I haven't checked carefully my working though.