Area Under The Projectile

#Area under the projectile A = ∫∫dxdy = ∫₀^R ∫₀^(y_max) dy dx = ∫₀^R (x tanθ - (g/2)x²) dx = (R²tanθ/2) - (gR³/6) = (4u⁴sin²θcos²θ/g) - (8u⁴sin³θcosθ/3g²) = (9.8u⁴sin²θcosθ/6g²) - (2u⁴sin³θ/6g²) If F is max, then dA/dθ = 0 → 3sin²θcosθ - sin³θ = 0 → tanθ = 3/2 → θ = 56.3° # Area under the projectile is max at θ = 56.3° # Range of the projectile is max at θ = 45° # Height of the projectile is max at θ = 90° Q. A particle of mass m and charge q is placed at origin with zero initial velocity at t=0. There exist uniform electric field E = Eî and uniform magnetic field B = Bk in the region. Find x(t), y(t), vx(t), vy(t). Solution: initial force (net) Fnet = qEî + q(vx)Bk Fnet = qEî as vx = 0 Equation of motion at time t: m(d²x/dt²) = qEx + q(dy/dt)B m(d²y/dt²) = qEy - q(dx/dt)B Solving the differential equations: m(d²x/dt²) = qE - q²B²/m * x m(d²y/dt²) = q²B²/m * y Second order differential equations: x'' + (q²B²/m²)x = qE/m y'' + (q²B²/m²)y = 0 Solutions: x = C1cos(ωt) + C2sin(ωt) + qE/(mω²) y = C3cos(ωt) + C4sin(ωt) From the initial conditions: at t = 0, x = 0, y = 0, dx/dt = 0, dy/dt = 0 we get: C1 = 0, C3 = 0, C2 = qE/(mω²), C4 = 0 Therefore: x = (qE/(mω²))(1 - cos(ωt)) y = (qE/(mω²))(sin(ωt)) What is total distance moved by the particle (ball) before coming to rest in previous question? Solution: distance till 1st collision = H if distance by 1st & 2nd collision = H/9, then distance = 9H/9 = H if distance by 2nd & 3rd collision = H/81, then distance = 9H/81 = H/9 distance between 1st and 2nd collision = H/9 distance between 2nd and 3rd collision = H/81 distance between 3rd and 4th collision = H/729 Total distance = H + H/9 + H/81 + H/729 + ... = H(1 + 1/9 + 1/81 + 1/729 + ...) = H(1 + 1/(1 - 1/9)) = 5H/4 → A freely falling particle, falling from a tower of height h covers a distance h/2 in the last second of its motion. The height of the tower is, nearly (g = 10 m/s²)   (a) 50 m (b) 55m (c) 45 m 58m Salution since it is mation under gravity : velacity after ½ distance; v₁ = √(gh/2) velacity after h distance; v₂ = √(gh) Since it gains velacity √(gh/2) from √(gh) is one second v - u = at √(gh) - √(gh/2) = g(1) √(gh) = √(gh/2) + g h = (√(gh/2) + g)² / g = (√(10h/2) + 10)² / 10 = (√(5h) + 10)² / 10 = (1.414√h + 10)² / 10 ≈ 58.34