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A long jumper leaves the ground at an angle of 30.0 to the horizontal and at a speed of 120m/s i. How far does he jump ? i. What is his velocity after 0.5s?​

User Tilo Mitra
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1 Answer

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(i) The horizontal distance traveled by the jumper 1,272.53 m.

(ii) The velocity of the jumper after 0.5 s is 117.6 m/s.

How to calculate the horizontal distance of the jumper?

(i) The horizontal distance traveled by the jumper is calculated by applying the following formula as shown below.

R = (u² sin (2θ) ) / g

where;

  • u is the initial velocity of the jumper
  • θ is the angle of projection
  • g is acceleration due to gravity

R = (120² x sin (2 x 30) ) / 9.8

R = 1,272.53 m

(ii) The velocity of the jumper after 0.5 s is calculated as follows;

v = u + gt

where;

  • v is the final velocity
  • u is the initial velocity

in vertical direction:

v = 120 sin(30) - 9.8 x 0.5

v = 55.1 m/s

in horizontal direction:

v = 120 m/s x cos(30)

v = 103.92 m/s

The resultant velocity is;

v = √ (55.1² + 103.92²)

v = 117.6 m/s

User Nishutosh Sharma
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