A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two times of flight is proportional to

Here t1=2usin⁡αg and t2=2usin⁡90∘−αg…(1)Now t1t2=4u2g2sin⁡αcos⁡αWe know thatR=u2sin⁡2αgFrom eqs. (1) and (2), we get t1t2=2Rg∝R

A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two times of flight is proportional to

Here t1=2usin⁡αg and t2=2usin⁡90∘−αg…(1)Now t1t2=4u2g2sin⁡αcos⁡αWe know thatR=u2sin⁡2αgFrom eqs. (1) and (2), we get t1t2=2Rg∝R

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A projectile can have the same range R for two angles of projection. If t₁ and t₂ be the times of flights in the two cases, then the product of the two times of flight is proportional to

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1/R

1/R²

R²

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Here $t_{1} = \frac{2 usin α}{g} and t_{2} = \frac{2 usin (90^{\circ} - α)}{g} \dots (1)$

Now $t_{1} t_{2} = \frac{4 u^{2}}{g^{2}} \sin αcos α$

We know that

$R = \frac{u^{2} \sin 2 α}{g}$

From eqs. (1) and (2), we get

$t_{1} t_{2} = \frac{2 R}{g} \propto R$

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A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two times of flight is proportional to

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A projectile can have the same range R for two angles of projection. If t₁ and t₂ be the times of flights in the two cases, then the product of the two times of flight is proportional to