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Physics 20 Online
OpenStudy (anonymous):

A golf ball is shot up into the air with a velocity of 37m/s at an angle of 35 degrees with the horizon a) how long will it take to hit the ground? b)how high will it go? c) how far away from the golfer will it land?

OpenStudy (anonymous):

This question is a projectile question, so the use of the kinematic equations will be most helpful. how much depth do you want or need help with on this question? I can be very specific or can just help in little parts where you are stuck.

OpenStudy (anonymous):

very specific please

OpenStudy (anonymous):

ok! are you familiar with the kinematic equations? there are three of them

OpenStudy (anonymous):

noo

OpenStudy (anonymous):

alright, well... we will use the kinematic equations, i'll assume you know them. first we need to convert the velocity into an x and y componetn of velocity like this.. |dw:1381808935269:dw|

OpenStudy (anonymous):

oh yeAh i know how to do that with the sin and cos. i have those values idk what to do with them though

OpenStudy (anonymous):

so to get the velocity component in the y direction we know that sin(35) = opposite/hypotenuse or sin(35)* 37 = 21.22 m/s in the y direction oh ok good so the next part is...

OpenStudy (anonymous):

to break up the golf ball up into two parts, we always break into two parts, becuase this enables to assign a value of zero to Velocity. like this: V = V + at 0 = V + at so what we are saying is the veloicyt is equal to zero when the projectile is at it's peak. it has zero velocity becuase it's no longer traveling up, nor is it traveling down at this instant make sense?

OpenStudy (anonymous):

yup it does make sense.

OpenStudy (anonymous):

alright, so a is equal to gravity, in our case its si units, so we use -9.81 m/s^2

OpenStudy (anonymous):

what equation do i plug it into

OpenStudy (anonymous):

the one i'm using above V = V + at 0 = V + at

OpenStudy (anonymous):

i'll draw a picture so you get the better idea..

OpenStudy (anonymous):

|dw:1381809367160:dw|

OpenStudy (anonymous):

its alright im following alomng so far

OpenStudy (anonymous):

so this is the first part - the ball moving from the ground to its peak, the second part is the ball at its peak to hitting the ground.

OpenStudy (anonymous):

okay i see it and we need to find the second half of the arc

OpenStudy (anonymous):

ok ok so question a says we need to find the time the ball is in the air. so we will first find the time the ball takes to reach it's peak, then find the time it takes to fall from it's peak

OpenStudy (anonymous):

so dont we find the time to when it reaches its peak and then square it? thats what my teacher said

OpenStudy (anonymous):

ya, the second half is what we need now. did the first part look like this.. 0 = 21.22 + -9.81*t ?

OpenStudy (anonymous):

haha, our responses are getting mixed up..

OpenStudy (anonymous):

yeah thats what it looked like and were on the same page now haha

OpenStudy (anonymous):

great, so do you understand the first part of the arc to get the time?

OpenStudy (anonymous):

is it 11.41?

OpenStudy (anonymous):

i solved this0 = 21.22 + -9.81*t

OpenStudy (anonymous):

that is the right equation... but incorrect time, try it again one more time solving it and if not, i'll help

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