solution: velocity of projection(u)=49m/s angle of projection=60degrees Maximum height(H)=H=(u2sin2Ɵ)/2g =(49x49xsin2(60)/2×9.8 =91.8m
solution: The particle projected with (u)=10m/s H=3.2M a=10m/s2 S= ut +(1/2)at2 3.2=10t-1/2x10xt2 3.2=10t-5t2 t=0.4s
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solution: a body falls freely from the tower ,it means it has initial velocity is zero. we have formula ,The total time taken is ‘t’ S= ut +(1/2)at2 Total height (H)=(1/2)g(t)2 …………. (1) For remaining 64% covered in (t-1) seconds (64/100)H=1/2g(t-1)2 ……………….(2) Equation (1)/Equation(2) 1/(64/100)=(t)2 /(t-1)2 (t/t-1)=10/8 t=5sec then, Total height (H)=(1/2)g(t)2 =(1/2)*10*5*5 H =125m
SOLUTION: Velocity at half of max = root 5/2
answer: A freely falling body moving with gravitational acceleration because it moves under Gravitational force.
ANSWER:M0L1T-2
solution: a body projected with velocity (u)=49m/s projected with an angle =60 degrees H=(u2sin2Ɵ)/2g H=(492sin260)/2g H=(49*49*3/4)/2*9.8 H=91.875m
solution: velocity of bullet after passing from plank ,its velocity decreases . initial velocity(u)=u final velocity(v)=initial velocity-1/20 of initial velocity v=u-u/20 v=19u/20 we have formula for one plank, (v)2-(u)2=2as, (19u/20)2-(u)2=2as 39u2/400=2as ……….(1) for n planks: then width of one plank will be ‘s’ width of n planks will be ns after passing from n planks …
