A long jumper leaves the ground at 45∘ above the horizontal and lands 7.1 m away. What is her "takeoff" speed v0?

Answer 1

Range = u²Sin (2theta) /g

as Sin2theta = 1 ,at theta = 45 , So Range is Max , As Sin 90 =1

Max range =u² /g

Max Range=7.1 m

u = initial speed =

So , u² = 7.1*g

= 7.1*9.8

u = √7.1*9.8=√(68.6) m/s = V₀= 8.3m/s