Question

The range of a projectile when fired at angle 75 to the horizontal is 0.5 km. what will be its range when fired at 45 with same speed..please explain properly

Answer 1

Hi

horizontal range =u^2sin2Q/g

sin150=1/2

swe get same value

Answer 2

The range of the projectile would be 1.0m when fired at 45°

Explanation:

$R_{1}$ = $\frac{u^{2} sin150}{g}$

= $\frac{u^{2} sin(180 -30) }{g}$

= $\frac{u^{2 sin30} }{g}}$

= $\frac{u^{2} }{2g}$ = 0.5m

$R_{2}$ = $\frac{u^{2} sin90 }{g}$

   = $\frac{u^{2} }{g}$

  = $2R_{1}$

  = 1.0m

Learn more: Range of projectile

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