Question

A body of mass m, projected vertically upwards with an initial velocity 'u' reaches a maximum
height 'h'. Another body of mass m, is projected along an inclined plane making an angle 30' with
the horizontal and with speed 'u'. The maximum distance travelled along the incline is​

Answer 1

Answer

• The maximum Distance is 2h

Explanation

Given

• A body has a mass m
• It is projected vertically upward
• Initial Velocity is 'u' & Maximum Height is 'h'
• Another body is projected with θ as 30°

To Find

• The maximum distance along the incline

Solution

✭ When thrown upward

→ v²-u² = 2as

→ v²-u² = 2 × (-g) × h

→ 0²-u² = -2gh

→ -u² = -2gh

→ u² = 2gh

→ h = u²/2g

✭ When Projected along the incline

→ v²-u² = 2as

→ v²-u² = 2 × (-g sinθ) × s

→ v²-u² = -2gsin(θ)s

→ 0²-u² = -2gsin(θ)s

→ -u² = -2gsin(θ)s

→ u² = 2gsin(θ)s

→ u²/2gsin(θ) = s

→ u²/2g × 1/sin(θ) = s

→ h × 1/sin(30°) = s

[sin 30° = ½]

→ h/½ = s

→ s = 2h

Answer 2

When thrown vertical, v^2 - u^2 = 2gh;v = 0 On inclined plane, v^2 - u^2 = 2gsintheta h1 = gh1 as theta = 30^o From above equations, h1 = 2h