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Grade: 11
        An object is projected with velocity V. If the range R of this object is double the maximum height H, then its range is1. V^2/g2. 3v^2/5g3. 4v^2/5g4. V^2/2g
2 years ago

Answers : (1)

Vikas TU
9499 Points
							
Range = V^2*sin(2*thetha)/g
and
H = V^2*(sin(thetha))^2/2g
given, R = 2H
Therefore,
V^2*sin(2*thetha)/g = V^2*(sin(thetha))^2/2g
=> 2sin(thetha)cos(thetha) = sin(thetha)^2/2 
on solving we get sinthetha = 4/root(17)
and the Range => 8V^/17g
 
2 years ago
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