you stand on the edge of a cliff of height (h) and throw astone as hard as you c
ID: 1737448 • Letter: Y
Question
you stand on the edge of a cliff of height (h) and throw astone as hard as you can. The equations of motion describingthe stone are x=(15.0 m/s)t and y=(13.5 m/s)t-(4.9 m/s^2)t^2 a) given the above equations of motion, where is thecoordinate system origin? b) What is the launch angle of the stone? c) How fast did you throw this stone?d) What is the height (h) of this cliff? you stand on the edge of a cliff of height (h) and throw astone as hard as you can. The equations of motion describingthe stone are x=(15.0 m/s)t and y=(13.5 m/s)t-(4.9 m/s^2)t^2 a) given the above equations of motion, where is thecoordinate system origin? b) What is the launch angle of the stone? c) How fast did you throw this stone?
d) What is the height (h) of this cliff?
Explanation / Answer
Part A) It is clear from these equations that the origin is whereyou threw the stone since at time=0; x=0 and y=0 (aka origin) andthe stone at t=0 is in your hand Part B) From the x equation the initial velocity in the horizontaldirection is 15m/s and the initial velocity in the y direction is13.5 m/s in the vertical direction. (the 4.9 is1/2*a=1/2*g=1/2*9.8=4.9). So to find the angle we know tan ()= y/x so tan()=13.5/15 =42.0 deg C) Total Speed= sqrt(horizontal speed^2+veretical speed^2) Total Speed= sqrt(15^2+13.5^2) Total Speed=20.2 m/s D) To find the height of cliff you need to know how long the stoneis in flight and you can plug that time into the y equation and itwill give you the negative of the height, also if it was given howfar the stone flew horizontally you could plug that into thex-direction and get the time that way and then plug that time intothe y-equation
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