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 $
