Projectile - Two dimensional motion - air resistance is neglected. Path followed is parabolic in nature called trajectory. * Let a projectile be projected with initial velocity u making an angle of x with the horizontal then its velocity can be resolved in two parts: i) Horizontal velocity= u cosx ii) Vertical velocity= usinx - The horizontal velocity remains constant where vertical velocity changes due to gravity. - Maximum height( Hmax)= (u^2 (sinx)^2)/2g - Time of flight (T)= (2usinx)/g - Horizontal range( R)= (u^2 sin2x)/g - For the maximum horizontal range the projectile should be projected at an angle of 45, R(max)= u^2/g. - Horizontal range is same for to angles x and 90-x. ( complementary angles) - H( max) = (gT^2)/8 - 4H(max) cot x= R - If R= H( max) then x= 75.96 degree - At maximum height, the velocity is perpendicular to the acceleration. - If to attain a certain height in T1 second and returns back to same point in T2 second then., H( max)= 1/2 g* T1*T2 a. Change in different from point of projection to the maximum height, E be the energy at initial point i) Change in speed= u- u cos x = 2u^2 (sin x/2)^2 ii) Change in velocity = u sin x iii) Change in KE= E (sinx)^2 iv) Change in PE= E (cosx)^2 v) change in direction = x a. Change in different from point of projection to the striking point, E be the energy at initial point i) Change in speed= 0 ii) Change in velocity= 2 u sinx iii) Change in momentum= 2mu sinx iv) Change in KE= 0 v) Change in PE= 0 vi)Change in direction= 2x * If the air resistance is not neglected then, R, H( max), speed at the string point is less...
Posted on: Sat, 18 Oct 2014 12:07:59 +0000
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