v2sin(2theta)=100g, vsin(theta)=5g,thus v2=25g2/sin2(theta), thus 50g2cot(theta)=100g, theta=arccot(1/5), v=10√26 m/sec
time of flight=2usin@/g=10........1
horizontal range=u^2sin2@ /g=100........2
dividing 1by2
usin2@/2sin@=10 or
ucos@=10....3
divide 1 by 3
we get tan@=5 ,cos@=1/sqrt26
on putting value of @ in 3 we get
u=10sqrt26
Let θ be the angle of projection (with the horizontal) and ‘u’ be the velocity of projection. Given time of flight is, t = 10 s.
Horizontal range, x = 100 m
=> u cosθ × 10 = 100
=> u cosθ = 10 …………………..(1)
=> u2 cos2θ = 100 ………………..(2)
Now, considering the vertical motion of the projectile,
v = u + at
=> -u sinθ = u sinθ – gt (since, initial vertical velocity = u sinθ, final vertical velocity after time t is –u sinθ)
=> u sinθ = gt/2 = (10 × 10)/2 = 50 ………………..(3)
=> u2 sin2θ = 2500 ………………………………….(4)
(2) + (4) => u2 = 2600
=> u = 50.99 m/s
(3)/(1) => tanθ = 5
=> θ = 78.690
In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?
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