Question

The range of a particle when launched at an angle of 15 degree with the horizontal is 1.5
km. What is the range of the projectile when launched at an angle of 45 degree to the horizontal?

Answer 1

Answer:

this is your correct answer mate

3KM

Answer 2

Answer:

The range of the projectile is 3 km.

Explanation:

The range of the projectile is given as,

$R=\frac{u^2sin2\theta}{g}$              (1)

Where,

R=range of the projectile

u=velocity of the projection

θ=angle of projection with respect to the ground

g=acceleration due to gravity=10m/s²

When θ=15°

$1.5=\frac{u^2sin2\times 15\textdegree}{g}=\frac{u^2sin30\textdegree}{g}$

$1.5=\frac{u^2}{2g}$               ($sin30\textdegree=\frac{1}{2}$)

$3=\frac{u^2}{g}$          (2)

When θ=45°

By using equation (1) we get;

$R_2=\frac{u^2sin2\times 45\textdegree}{g}=\frac{u^2sin\times 90\textdegree}{g}$        ($sin90\textdegree=1$)

$R_2=\frac{u^2}{g}$            (3)

By taking the ratio of equation (2) and equation (3) we get;

$\frac{3}{R_2}=\frac{\frac{u^2}{g}}{\frac{u^2}{g}}=1$

$R_2=3km$

Hence, the range of the projectile when launched at an angle of 45 degrees to the horizontal is 3km.