Asked by Justine

You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 22.3 m/s at an angle of 55.0° above the horizontal. If it hits the top of the wall at a speed of 15.4 m/s, how high is the wall?
Answered by uoit
I see you! I see what you are doing my fellow uoit physics mate :D
Answered by UOIT studen
hahaa another module question that im stuck on, ayyy
Answered by UOIT STUDENT
use the vf^2=vi^2 + 2*a*sin(55)*d
solve for d
Answered by UOIT Student
LOL mastery modules smh
Answered by another uoit kid
ayyy 3 years later and we're still suffering
Answered by UOIT Comrade
Struggling through the module right now, figured out how to do this question: I got y'all:
Find x and y initial velocity components:
Viy=22.3sin55 = 18.3 m/s. Vix=22.3cos55= 12.8 m/s
Since horizontal velocity is constant, you can use the Vf^2 equation to solve for Vyf.
15.4^2=12.8^2+Vyf^2. Power through and rearrange to solve Vyf.
237.16- 163.84= Vyf^2
73.32=Vyf^2, so Vyf= 8.56 m/s
Now that you got this, you can use the one of the Kin equations in the y component to solve for the change in position (change in Y, and therefore, height).
Vfy^2=Viy^2 + 2a(delta y). Keep in mind ay is -9.8
8.56^2 - 18.3^2 = 2(-9.8)(delta y)
73.27 - 334.9 = -19.8 (delta y)
-261.63 = -19.8 (delta y)
-261.63/-19.8 = y (since initial y was 0).
13.2= y. So the castle is 13.2 metres high. Hope it helped! ~First year mech engineering/suffering student!
Answered by OTU trooper
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Answered by Evana
Ayyyy UOIT gang
Answered by another another OTU student
2020 in a global pandemic and we are still suffering, thanks tho
Answered by Another Another Another OTU Student
2020 is still being crummy, so thanks from me too.
Answered by OTU Cadet 2020
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Answered by bro uoit gang
uoit student reporting
Answered by O tech student
another o tech student here
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