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A ball is projected from a point O on the edge of a vertical cliff. The horizontal and vertically upward components of the initial velocity are 7ms-1 and 21ms-1 respectively. At the time t seconds after the projection the ball is at the point (x,y) referred to horizontal and vertically upward axes through O. Air resistance may be neglected. (i) Express x and y in terms of t, and hence show that y= 3x- 0.1x2The ball hits the sea at a point which is 25m below the level of O.(ii) Find the horizontal distance between the cliff and the point where the ball hits the sea.
(i) x=7t(horizontal velocity always remain constant so the formula used is s=vt) y=21t-0.5gt2(s=ut+0.5at2) t=(x/7) y=21t-0.5gt2 y=21(x/7) -0.5g(x/7) 2 y=3x-0.1x2(ii) y=-25(because the question said y is vertically upward axes through O the ball hits below O so negative 25) 3x-0.1x2=-25 0.1x2-3x-25=0 solving this equation we get x=36.8m
Have You Completed yous A Levels?
no,still doing my A2.
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Can u help with the second subpart too plz? Y do they use F = T cos (sin-1 0.6)
iVE quoted it.
But whats the ORIGINAL question ? All you've quoted is the answers Nid gave you - I cant find your question anywhere.
Click on the "Quote from...." mate.
I did that. I'm not dumb you know.... I figured out you were talking about Dania's question.This whole time I was looking for a question posted by YOU.
I know that the Friction force is stopping the ring to slip and opposing T too.I think it should be F = T + Wsin53.1 (weight component)But the answer seems to ignore the W, and instead give the equation F = Tsin53.1
W acts perpendicular to the ring...............hence it has no component!