An athlete executing a long jump leaves the ground at a 20° angle and travels 8.30 m. ?

Question

What was the takeoff speed(m/s)?If this speed were increased by just 4.0%, how much longer would the jump be(m)?

Answer 1

Sy=Uyt+0.5gt^2

0=Usin20t+0.5(-9.8)t^2

4.9t^2=Usin20t

4.9t=Usin20

t=Usin20/4.9

Sx=Uxt

8.30=Ucos20t

8.30=Ucos20(Usin20/4.9)

8.30=U(0.93)(U0.34/4.9)

8.30=0.064U^2

U^2=129.68

U=11.38 m/s

if increase 4% U=1.04*11.38=11.83

t=2Usinz/g = 2(11.83)sin20 / 9.8 = 0.82 s

Sx=Uxt=Ucos20t=11.83(0.93)(0.82) = 9.02 m

So,distance longer =9.02 - 8.30 = 0.72 m

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Answer 2

Angle of initial velocity made with horizontal = theta = 20 deg

Range R = 8.30 m

Initial velocity u = ?

R = u^2 * sin(2*theta)/g----------(1)

Or, u^2 = gR/sin(2*theta) = 9.8 * 8.3/sin 40 deg m/s = 126.54 m/s

If the speed were increased by 4.0%, then new speed = u + 4.0% of u

= u(1 + 4/100) = 1.04 u

Therefore, from (1)

new value of R = R * 1.04^2 = 8.30 * 1.04^2 = 8.98 m

So, the jump would be longer by 8.98 - 8.30 = 0.68 m

Answer 3

The working formula is

R = (V^2)(sin 2A)/g

where

R = range = maximum horizontal distance = 8.3 meters (given)

V = initial velocity

A = 20 degrees

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

8.3 = V^2(sin 2*20)/9.8

Solving for "V",

V = 11.25 m/sec.

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If the above speed were increased by 4%, then V = 11.7 m/sec and the corresponding range will be

R = (11.7)^2(sin 2*20)/9.8

R = 8.98 meters

Hence, the jump will be 8.98 - 8.30 = 0.68 m longer.