a bus travels at 25m/s and attempts to jump over a space between two parts of an unfinished bridge.the gap between the two points is 20m and the two are the same elevation.ehat minimum angle must the bus take off in order to make jump?
can u help me
jesus this is an annoying problem lol I hate angle problems!
:-(
too bad it doesn't give you a height that it reaches
they gave me x=20m and v=25m/s
i got answer as 9 degree
is that right
sec trying to work this out in my head kinetics has been a while
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units are in meters so we can use gravity constant g =9.8 m/s^2 we know the distance that needs to be cleared is 20 meters
lets start by looking at the x direction no acceleration only velocity distance travelled will be 20m we can use x=vt 20=(25 cos(theta))* t
ok
now lets look at the y position acceleration is g=-9.8m/s^2 velocity is v=25 sin(theta) the 2 points, where the bus leaves and lands are the same elevation/height we can call that x0=0 and y=0 at time t so we use y=1/2at^2+vt+c 0=-1/2(9.8)t^2 +25 sin(theta)
0=-1/2(9.8)t^2 +25 sin(theta)t
Yeah thats exactly what I was doing but my internet cut out lol
now we have a system of 2 equations with 2 unknowns, suggestion would be to solve for t and substitute
@completeidiot i got angle as 9
excatly i did that one i am not sure about algebra
i didnt actually solve the problem and i dont want to too much work
I got to 4.9 (20/(25cosx))^2 + 20 tanx= 0 then I got side tracked doing bio
suggestion- if you have a graphing calculator ti-84 graph the equation and find the root using the zero function
i know that but how do u find angle i can find time
x direction 20=(25 cos(theta))* t t=20/(25 cos(theta))
so how do u graph this
how do u put theta in it
theta =x
in calcu;ator
ok i will try
minimum angle of 9.138 degrees
so am i right?
well 9.138 > 9 yes, if you just rounded to the ones place technically no, if you actually attempted doing this, the bus would hit the cliff, everyone would die and it would fall to its doom or blow up
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