Question
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?
Answers
MathMate
Equate KE and PE to get
mgh = (m/2)vy²
=>
<b>vy=√(2gh)</b>
Initial velocity
v=vy/sin(θ)
Substitute v in range equation
<b>range=v²sin(2θ)/g</b>
= (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
mgh = (m/2)vy²
=>
<b>vy=√(2gh)</b>
Initial velocity
v=vy/sin(θ)
Substitute v in range equation
<b>range=v²sin(2θ)/g</b>
= (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