An athlete jumps at an angle of 300
with a maxim um speed of 9.4 m/s.
An athlete jumps at an angle of 300
with a maxim um speed of 9.4 m/s.
b) Obtain an expression to calculate the horizontal range covered by the athlete.
c) Find the range covered by him in the above jump. Suggest the angle by which the
athlete can attain the maximum
range.
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Answer:
(b) R = ucosθ. T
= ucosθ × 2usinθ/g
= u² sin2θ/ g
= (9.4)² sin(2 × 30°)/ 10
= 88.36 × root 3 / 2 × 10
= 7.652 m.
(c) The equation of motion says:
v=u+at
Considering only the vertical component of his jump we can get the time it takes to reach his maximum height from:
0=uy-gt
usine-uy which is the vertical component of his initial velocity u.
..0-usine-gt
..t=usineg
Since 0-23 this becomes:
t-4sin239.8=0.159s
The 2nd part of his journey as he falls to earth is a mirror image of the first part so:
Total time =2×0.159=0.32s
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