Asked by Eddie
Janet jumps off a high-diving platform with a horizontal velocity of 3.5 m/s and lands in the water 1.7 s later. How high is the platform, and how far from the base of the platform does she land?
Answers
Answered by
Tony
When finding how high the platform is the horizontal velocity isn't important.
S= 1/2at^2 where S is the displacement and a is the acceleration due to gravity (9.8)
so
S=(1/2) (9.8) 1.7^2
S=4.9*1.7^2
S=14.161 meters
When finding how far from the base Janet lands use the formula:
D=Ut where D= the distance from the platform, U= the horizontal velocity, and t= time
So:
D=Ut
D=3.5*(1.7)
D=5.95
Simple isn't it?
S= 1/2at^2 where S is the displacement and a is the acceleration due to gravity (9.8)
so
S=(1/2) (9.8) 1.7^2
S=4.9*1.7^2
S=14.161 meters
When finding how far from the base Janet lands use the formula:
D=Ut where D= the distance from the platform, U= the horizontal velocity, and t= time
So:
D=Ut
D=3.5*(1.7)
D=5.95
Simple isn't it?
Answered by
Kush
Janet jumps off a high diving platform with a horizontal velocity of 2.64 m/s and lands in the water 2.8 s later. How high is the platform? The acceleration of gravity is 9.8^2 m/s. Answer in units of m
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