- #1
vysero
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A stone is projected at a cliff of height h with an initial speed of 42m/s directed at angle=60 above the horizontal. The stone strikes at A, 5.5s after launching. Find A) the height h of the cliff, B) the speed of the stone just before impact at A, and C) the maximum height H reached above the ground.
ok so,
h= Vo(sin(60))t-1/2g(t)^2
so i have my answer for a which came out to be 51.8m
Here is where I am struggling with problem A. I only understand a small portion of why: h= Vo(sin(60))t-1/2g(t)^2
For instance why is Vo = Vo(sin(60)) ok so sin is opp/hyp the angle is 60 that gives me the length of opposite side so why is that not the answer for C?
I am obviously completely confused... help me...
ok so,
h= Vo(sin(60))t-1/2g(t)^2
so i have my answer for a which came out to be 51.8m
Here is where I am struggling with problem A. I only understand a small portion of why: h= Vo(sin(60))t-1/2g(t)^2
For instance why is Vo = Vo(sin(60)) ok so sin is opp/hyp the angle is 60 that gives me the length of opposite side so why is that not the answer for C?
I am obviously completely confused... help me...