Question

The range of the particle which is projected at an angle of ${15}^o$ is $1.5km$. What will be the range for an angle of projection ${45}^o$?

• A. $0.5km$
• B. $1.5km$
• C. $2.5km$
• D. $3km$

Answer 1

The correct option is D $3km$

Range $R=u^{2}Sin2\theta /g$

for angle $\theta ={15}^0$ we have range $R_1=u^{2}Sin{30}^0/g=1.5$ put $g=10$

we get $u^2=30$

Now the range for angle ${45}^0$ will be $R_2=u^{2}Sin{90}^0/g=u^2/g=30/10=3Km$

Option D is correct.