Kinematics Question 690

Question: The range of a particle when launched at an angle of $ 15^{o} $ with the horizontal it’s 1.5 km. What is the range of the projectile when launched at an angle of $ 45^{o} $ to the horizontal

Options:

A) 1.5 km

B) 3.0 km

C) 6.0 km

D) 0.75 km

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ {R _{15{}^\circ }}=\frac{u^{2}\sin (2\times 15{}^\circ )}{g} $

$ =\frac{u^{2}}{2g}=1.5km $

$ {R _{45{}^\circ }}=\frac{u^{2}\sin (2\times 45{}^\circ )}{g}=\frac{u^{2}}{g}=1.5\times 2=3km $



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