answer me 2nd its urgent
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here the maximum height of both projectiles is same but range of 2nd is more than 1st
=>max height H = U^2*(sin(theta))^2/(2g)
H is constant thus usin(theta) is constant
since max height is same velocity is inversely proportional to sin(theta) thus for higher theta velocity is less. since angle of 1st projectile is more than 2nd V1 < V2
=>Now time of flight is T = 2usin(theta)/g = constant
Thus T1=T2
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=>max height H = U^2*(sin(theta))^2/(2g)
H is constant thus usin(theta) is constant
since max height is same velocity is inversely proportional to sin(theta) thus for higher theta velocity is less. since angle of 1st projectile is more than 2nd V1 < V2
=>Now time of flight is T = 2usin(theta)/g = constant
Thus T1=T2
check my bio and follow me for more help
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