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Applied Mechanics - Engineering

Q1:

If the horizontal range is 2.5 times the greatest height, the angle of projection of the projectile, is

A 57°

B 58°

C 59°

D 60°

ANS:B - 58°

Horizontal range,R= (u^2sin(2$))/g........(eq1)
Maximum height,H=(u^2sin^2($))/2g........(eq2)

Here,
Horizontal range= 2.5*Maximum height
R=2.5*H.....(eq3)

Substitute eq 1 & eq 2 in eq3

Sin(2$)=(2.5sin^2($))/2
2sin($)cos($)= (2.5sin^2($))/2
tan($)=1.6
$=57.99°=58°