OpenStudy (ksaimouli):

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?

5 years ago
OpenStudy (ksaimouli):

can u help me

5 years ago
OpenStudy (anonymous):

jesus this is an annoying problem lol I hate angle problems!

5 years ago
OpenStudy (ksaimouli):

:-(

5 years ago
OpenStudy (anonymous):

too bad it doesn't give you a height that it reaches

5 years ago
OpenStudy (ksaimouli):

they gave me x=20m and v=25m/s

5 years ago
OpenStudy (ksaimouli):

i got answer as 9 degree

5 years ago
OpenStudy (ksaimouli):

is that right

5 years ago
OpenStudy (anonymous):

sec trying to work this out in my head kinetics has been a while

5 years ago
OpenStudy (anonymous):

|dw:1348344084097:dw|

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (ksaimouli):

ok

5 years ago
OpenStudy (anonymous):

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)

5 years ago
OpenStudy (anonymous):

0=-1/2(9.8)t^2 +25 sin(theta)t

5 years ago
OpenStudy (anonymous):

Yeah thats exactly what I was doing but my internet cut out lol

5 years ago
OpenStudy (anonymous):

now we have a system of 2 equations with 2 unknowns, suggestion would be to solve for t and substitute

5 years ago
OpenStudy (ksaimouli):

@completeidiot i got angle as 9

5 years ago
OpenStudy (ksaimouli):

excatly i did that one i am not sure about algebra

5 years ago
OpenStudy (anonymous):

i didnt actually solve the problem and i dont want to too much work

5 years ago
OpenStudy (anonymous):

I got to 4.9 (20/(25cosx))^2 + 20 tanx= 0 then I got side tracked doing bio

5 years ago
OpenStudy (anonymous):

suggestion- if you have a graphing calculator ti-84 graph the equation and find the root using the zero function

5 years ago
OpenStudy (ksaimouli):

i know that but how do u find angle i can find time

5 years ago
OpenStudy (anonymous):

x direction 20=(25 cos(theta))* t t=20/(25 cos(theta))

5 years ago
OpenStudy (ksaimouli):

so how do u graph this

5 years ago
OpenStudy (ksaimouli):

how do u put theta in it

5 years ago
OpenStudy (anonymous):

theta =x

5 years ago
OpenStudy (ksaimouli):

in calcu;ator

5 years ago
OpenStudy (ksaimouli):

ok i will try

5 years ago
OpenStudy (anonymous):

minimum angle of 9.138 degrees

5 years ago
OpenStudy (ksaimouli):

so am i right?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (ksaimouli):

|dw:1348345211086:dw|

5 years ago