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Question: The range of a projectile which is launched at an angle of 15<sup>0</sup> with the horizontal is 1.5...

The range of a projectile which is launched at an angle of 150 with the horizontal is 1.5 km. What is the range of the projectile if it is projected at an angle 45° to the horizontal?

A

1.5 km

B

3 km

C

0.75 km

D

6 km

Answer

3 km

Explanation

Solution

R µ sin 2q

R1R2\frac { \mathrm { R } _ { 1 } } { \mathrm { R } _ { 2 } } = sin⁡2(15)sin⁡2(45)\frac { \sin 2 ( 15 ) } { \sin 2 ( 45 ) }