Ive tried and tried but i can\'t figure this out, over an hour wasted and nothin
ID: 2166716 • Letter: I
Question
Ive tried and tried but i can't figure this out, over an hour wasted and nothing gained. please help explain.A passenger on a moving train tosses a coin with an initial velocity of (-2.25 m/s)x unit vector + 4.92 m/s)y unit vector (from the point of view of the passenger). The train's velocity relative to the ground is (11.9m/s)x unit vector.
(a) What is the minimum speed of the coin relative to the ground during its flight?
At what part in the coin's flight does this minimum speed occur?
choose one from the following
1. immediately after being tossed
2. at some point as it is rising
3. at the top of its motion
4. at some point at it is falling
(b) Find the initial speed and direction of the coin as seen by an observer on the ground.
( ) m/s
( ) degrees (counterclockwise from the +x axis
(c) Use the expression for ymax derived in Example 4-7 to calculate the maximum height of the coin, as seen by an observer on the ground.
( )m
(d) Repeat part (c) from the point of view of the passenger.
( )m
Explanation / Answer
a) 13.5 m/s. At the top of the coins flight it has a vertical velocity of 0 and a horizontal velocity of 13.5 m/s b) Initial speed is (3.82^2+13.5^2)^.5 (Too lazy to open excel) initial direction is arctan(3.83/13.5) c) I don't have access to example 4-7 d) solve 3.82=9.8*t for the time it will take the coin to reach it's highest point x=.5*9.8*t simplified x=.5*9.8*(3.82/9.8) Total distance above the follow x+distance above the follw that he threw the coin from
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