Question

A particle is projected at the angle 37 with the incline plabe in upward direction with speed 10m/s. the angle of incline plane is 53. then the maximum distance from the incline plane attained by the particle will be

Answer 1

Answer:

The maximum distance from the incline plane attained by the particle will be 3.07 m.

Explanation:

Given that,

Projected angle = 37°

Speed = 10 m/s

Angle of incline plane = 53°

The maximum height attained by the particle from inclined plane

Formula of maximum height is defined as:

$H = \dfrac{u^2 \sin^2\theta}{2g\cos\alpha}$

Put the value into the formula

$H = \dfrac{(10)^2\sin^2(37^{\circ})}{2\times9.8\cos53^{\circ}}$

$H = \dfrac{100\times0.3621}{2\times9.8\times0.6018}$

$H = 3.07\ m$

Hence, The maximum distance from the incline plane attained by the particle will be 3.07 m.

Answer 2

Answer:

     Hmax = 3m

Explanation:

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