The maximum range of a projectile is 500 m. If the particle is thrown up a plane is inclined at an angle of 30o with the same speed, the distance covered by it along the inclined plane will be:

Question

The maximum range of a projectile is 500 m. If the particle is thrown up a plane is inclined at an angle of 30o with the same speed, the distance covered by it along the inclined plane will be: (A) 250 m (B) 500 m (C) 750 m (D) 100 m

Tardigrade Admin · Accepted Answer

For the maximum range, θ=45° R=(u2 sin 2θ/g)=(u2/g) sin 90°=(u2/g) or 500=(u2/g) The distance covered along the inclined plane can be obtained using the equation v2 -u2 =2as or 0-u2 =2(-g sin 30°)s or s=(u2/g) =500 m

Q. The maximum range of a projectile is 500 m. If the particle is thrown up a plane is inclined at an angle of 30o with the same speed, the distance covered by it along the inclined plane will be:

250 m

500 m

750 m

100 m

For the maximum range, θ=45∘ R=gu2sin2θ​=gu2​ sin 90∘=gu2​ or 500=gu2​ The distance covered along the inclined plane can be obtained using the equation v2−u2=2as or 0−u2=2(−gsin30∘)s or s=gu2​=500m

Q. The maximum range of a projectile is $500$ m. If the particle is thrown up a plane is inclined at an angle of $3 0^{o}$ with the same speed, the distance covered by it along the inclined plane will be: