MIRROR FORMULA The relation between object distance(u), image distance(v) and the focal length (f) of the mirror is known as MIRROR FORMULA. Consider a concave mirror MM' (of small aperture). An object AB of height h is placed on the left of the mirror beyond its center of curvature so that the image formed is real, inverted and diminished in size(height = h'). Here : Object distance = PB = -u Image distance = PB'= -v Focal length = PF = -f Radius of curvature = PC = -R As shown in Figure, Δ ABP ∼ Δ A'B'P. ∴ A'B'/AB = PB'/PB ∴ A'B'/AB = (-v)/(-u) ∴ A'B'/AB = v/u.............(I) Similarly, as shown in the figure, Δ ABC ∼ Δ A'B'C. ∴ A'B'/AB = CB'/CB ∴ A'B'/AB = (PC-PB')/(PB-PC)........[as CB'=(PC-PB') and CB=(PB-PC)] ∴ A'B'/AB = [-R - (-v)]/[-u - (-R)] ∴ A'B'/AB = (-R + v)/(-u + R)........(II) By (I) & (II), v/u = (-R + v)/(-u + R) ∴ -...
1) A bullet blasts from the barrel of a gun upward in the vertical direction with an initial speed of 700 m/s. Find the maximum altitude reached by this bullet and the time needed to reach it.
ReplyDeleteInitial speed u = 700 m/s Final speed v =0m/s (at the maximum altitude) , a = -10m/s2
t = ( v-u)/0 =( 0 – 700)/-10 = 70m/s
v2 = u2 + 2as
v2 - u2)/ 2a =s
0 – (700)2 /2x-10
= - 490000/-20 = 24500m
2) A mouse is dropped from an eagle's claws starting at an altitude of 150 meters. What distance does it fall in the third second after it is dropped?
S = ut + ½ at2
0x3 + ½ x 10 x3x3
= 45m
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