Question

# The range of projectile when launched at an angle of $$15^{\circ}$$ with the horizontal is $$1.5\ km$$. What is the range of the projectile when launched at an angle $$45^{\circ}$$ to the horizontal

A
1.5 km
B
3.0 km
C
6.3 km
D
0.75 km

Solution

## The correct option is C $$3.0\ km$$ Here in first case, $$R_{1} = 1.5 , \theta = 15$$ $$R_{1} = \dfrac{u^{2}\sin (2\times 15) }{g}\Rightarrow 1.5 = \dfrac{u^{2}}{2g}$$  $$\Rightarrow \dfrac{u^{2}}{g} = 3$$ Now, for second case, $$R_{2} = \dfrac{u^{2}\sin (2\times45) }{g}\Rightarrow R_{2} = \dfrac{u^{2}}{2g} = 3 km$$ Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More