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khanitha
A piece of marble is projected from earth's surface with velocity of 50 m/s. 2 seconds later, it just clears a wall 5 m high. What is the angle of projection
Maybe you could make a triangle out of it so 2 seconds it barely clears the wall so the hyp is 100m and the opposite side from the angle is 5m and use inverse sin to find the angle so oppo/hyp 5/100, I got 2.87 degrees that seems a little small though...
|dw:1333208034298:dw| heres a picture
apparently its 60 degrees says another site
@awstinf u totally ignord gravity acting downward(it gives the body accelerated motion)not uniform
so wat u do is in horizontal direction,the cmponent of velocity is 50cos@ in time 2 seconds it will cover a distance 50cos@*2 |dw:1333208698171:dw| tan@=sin@/cos@=100/50cos@ solve it
shouldnt there be a distance part to this?
|dw:1333208865379:dw| sry i made a mistake in the diagram
lol, i spose we are doing: 2(50) cos(a) to determine a distance from the wall :)
\[h(t)=-\frac{g}{2}t^2+V_o sin(a)\ t+H_o\] \[h(2)=5=-9.8+2(50)sin(a)+0\] \[5+9.8=100sin(a)\] \[\frac{14.8}{100}=sin(a)=.1480\] \[a=sin^{-1}(.1480)\] maybe
something has caught me! i am reading all the question wrong1 |dw:1333209509913:dw| use 5m instead of 100m that gives tan@=sin@/cos@=5/100cos@ sin@=1/20=0.05
amistre ur approach is absolutely right check ur 9.8t^2 part u gobbled up the t^2
:) yeah, i seen something was amiss, but couldnt place it. thanx
Hi use this formula\[y=(-1/2)g t^2+v _{0}\sin \theta+y _{0}\] if you take y=5,g=9.81,t=2,v0=50,y0=0 theta yields29.49 degrees