Answer to Question #82640 in Physics for loai

Question #82640

A long jumper leaves the ground at 45∘
above the horizontal and lands 8.2 m
away. What is her "takeoff" speed
v0

Expert's answer

A jumper horizontal distance (range)

R=(v_0^2 sin⁡2θ)/g

So, a jumper initial speed

v_0=√(gR/sin⁡2θ )=√((9.8×8.2)/sin⁡〖90°〗 )=9.0 m/s

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