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⏺ A projectile is projected with velocity 20 m/s , such that range is 4 times maximum then calculate
a) R
b) H
c) T
d) theta
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Answers
Explanation:
Velocity (u) = 20m/s^2
According to the question,
Range = 4× Maximum Height
u^2 sin 2A/g = 4× u^2 Sin^2 A/2g
Where A is the angel of projection
Sin2A = 4× sin^2 A/2
=> 2 SinA × CosA =4× sin^2 A /2
=> CosA = SinA
=> sinA/CosA = 1
=> tanA = 1
=> tanA =tan45°
=> A =45°
(a) R = u^2 Sin2A /g
= 20^2 × Sin2×45°/10
= 400×Sin90/10
= 40 × 1
= 40 m
(b) H = u^2 sin^2 A/2g
= 20^2 × sin^2 45°/2×10
= 400 ×1/2 /20
= 400/40
= 10 m. ANS
(C) T = 2u Sin A/g
= 2× 20 ×Sin 45°/10
= 40/10 × root2
= 4/ root2 sec
(D). Theta is 45° As we have taken angle of projection AS " A"
Answer:
400 / g ; 100 / g ; 20 / 9 ; 45°
Based on the value of g numeric values of these can be determined.
Explanation:
From the properties of projectile motion( free falling body ) :
Height is given by { u²sin² A / 2g } ; range( horizontal ) is given by { u²sin2A / g } ; total time is given by ( 2u.sinA / g )
*where A is the angle of projection ; symbols have their usual meaning.
Here,
= > Range = 4 Height
= > u²sin2A / g = 4( u²sin²A / 2g )
= > sin2A = 2sin²A
= > 2sinAcosA = 2sin²A { sin2A = 2sinAcosA }
= > sinA = cosA
= > sinA = sin( 90 - A ) { cosB = sin(90-B)
= > A = 90 - A
= > 45°
Thus,
angle of projection is 45°( theta ).
Range R :
= > u²sin2A / g
= > ( 20 )^2 x sin2( 45° ) / g
= > 400 sin90° / g
= > 400 x 1 / g
= > 400 / g m
Height H :
= > u²sin²A / 2g
= > ( 20 )^2 x sin²( 45° ) / 2g
= > 400 x ( 1 / √2 )² / 2g
= > 400 x 1 / 4g
= > 100 / g m
Time T :
= > 2u.sinA / g
= > 2 x 20 x sin45° / g
= > 2 x 20 x 1 / 2g
= > 20 / g sec