Question

Q7) A man thrown a ball to maximum horizontal distance of 8 meters. Calculate the
maximum height reached.​

Answer 1

Answer:

u²{1 + √(1+160/u²)} ÷40

Explanation:

8 = u²sin2∝/g = u²sin2∝/10 , sin2∝ = 80/u² , sin∝cos∝ = 40/u²

sin²∝cos²∝ = 40/u² . sin²∝ + cos²∝ = 1 , let sin²∝ = s²

s²( 1 - s²) = 40/u²

s∧4 - s² + 40/u² = 0

s² = {1 ± √(1+160/u²)} ÷2 =  {1 + √(1+160/u²)} ÷2

max height reached = u²sin²∝/2g  =  u²{1 + √(1+160/u²)} ÷40