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?
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!
ayyy 3 years later and we're still suffering
LOL mastery modules smh
2020 in a global pandemic and we are still suffering, thanks tho
I see you! I see what you are doing my fellow uoit physics mate :D
hahaa another module question that im stuck on, ayyy
I respect your decision in sharing this information
use the vf^2=vi^2 + 2*a*sin(55)*d
solve for d