The question An athlete executing a long jump leaves the ground at a 35° angle and travels 6.30 m.?
I DID MY WORK IN FINDING THE TAKEOFF SPEED.
R = v0^2 sin(2theta)/g for projectiles landing at the same height they launch
here, theta =35 and R=6.3m
we have:
v0^2=g R/sin( 2 theta) = 9.8x6.3/sin(70) = 65.7
v0=sqrt[65.7]=8.1m/s------->is the takeoff speed
my question is If this speed were increased by just 6.0%, how much longer would the jump be? ANSWER SHOULD BE IN m.
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hi kb !
good to see you do part of the q before posting !
the calculation is now very simple.
from the formula, it is clear that
R varies directly as the SQUARE of the launch velocity,
(other factors remaining the same)
so if the launch velocity increases by 6% to become 1.06 of its initial value,
range will increase to 1.06^2 of initial value,
= 1.06^2*6.3
= 7.08 m to 2 dp
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