An archer stands a horizontal distance d= 55 m away from a tree sees an apple hanging from the tree at h= 12 m above the ground. The archer chooses an arrow and prepares to shoot. The arrow is initially 1.5 m above the ground. Just as the archer shoots the arrow with a speed of 80 m/s , the apple breaks off and falls straight down. A person of height 2.0 m is standing directly underneath the apple. The arrow pierced the apple. Ignore air resistance, and use g=9.81 m/s2 for the acceleration of gravity. What angle did the archer aim the arrow? enter your answer in degrees
Hey this is from Edx... :P.. plz don't answer this.. you are not supposed to get outside help .. code of conduct dude..
plz i need help
i ll help you.. tell me what part don't u get?!
how to find the angle and at what height should i put ?
i got the distance
u do remember that monkey thing right.. if you aim right at the monkey and while u fire the monkey falls, then it ll hit for sure so here its given that the arrow hits the apple.. so doesn't it mean, that he aimed right at it?!
yup....so applying rangle formula right? R=
no.. draw it .. u ll get it!
i gt 12 and its wrong
start by drawing it
did that
solve for the time of the apple falling
t=d/v_0cos(thetha)
yeah, you have all those values right?
so get a number for t and sub it into the arrow equations
i got the arrow pierces apple at a height of 7.97m above persons head
tell me how, if you have got that u can use r=v^2sin2theta/g
Hi super scientist, how did you get the distance I don't want the answer obviously but some guide I am stuck in this one!!
how did u get 7.97 huh?
my thoughts are: i would determine an interval of time that it takes the apple to fall from 12m to 2m the arrow need to cover 55 m at a speed of: 80t cos \(\alpha\)
\[0 <=~t~<= \sqrt{\frac{20}{g}}\]
i determined that by \[-\frac12gt^2+10=0\] \[-\frac12gt^2=-10\] \[gt^2=20\] \[t^2=\frac{20}{g}\] \[t=\sqrt{\frac{20}{g}}\]
how did you get a 20 up there
got it
we can then define a range for the angle knowing the distance and initial velocity of the arrow:\[80t~\cos(\alpha)=55\] \[\cos(\alpha)=\frac{55}{80t}\] \[\alpha=\cos^{-1}\left(\frac{55}{80t}\right)\] t = 0 is out of the question :) any others we can omit?
i get a range of angles from 0 to 70 i believe
lol, i forgot to sqrt my 20/g
0 to 61
i would then try to narrow it down by determining how far the arrow would "drop" perse
|dw:1380114047701:dw| something like this
Join our real-time social learning platform and learn together with your friends!