a long jumper takes off at an angle of 20 degrees with the horizontal and reaches a maximum height of 0.55 meters at mid- flight. how far did she jump?

Answer 1

Equate KE and PE to get
mgh = (m/2)vy²
=>
vy=√(2gh)

Initial velocity
v=vy/sin(θ)

Substitute v in range equation
range=v²sin(2θ)/g
= (vy²/sin²(θ))*sin(2θ)/g
= (vy²*2sin(θ)cos(θ)/(g*sin²(θ)
= (2gh*2cos*theta;/(g*sin(θ)
= 4h/tan(θ)
= 4*0.55/tan(20°)
= 6.0 m