Question

An athlete performing a long jump leaves the ground at a 32.9 ∘ angle and lands 7.78 m away.part a )what was the takeoff speed ?

part b) if this speed were increased by just 8.0% , how much longer would the jump be ?

Answer 1

a)Vo = 9.58 m/s.

b)D= 9.04

The jump would be 1.26 m Longer.

Step-by-step explanation:

Write down the values given in the question

The athlete performing a long jump leaves the ground at 32.9° angle.

The athlete lands 7.78 m away.

a)

D = Vo^2*sin(2A)/g = 7.78 m.

Vo^2*sin56 / 9.8 =7.78

0.08460*Vo^2 = 7.78.

Vo^2 = 7.78 / 0.08460 = 91.962

Vo = 9.58 m/s.

b)

Vo = 1.08 * 9.58 = 10.34 m/s.

D = (10.34)^2*sin56/9.8

=106.9156* 0.0845= 9.04.

D= 9.04

D2 - D1 = 9.04 - 7.78 = 1.26 m Longer.

Answer 2

Step-by-step explanation:

Given

Inclination
`\theta =32.9^(\circ)`

Distance of landing point
`R=7.78\ m`

Considering athlete to be an Projectile

range of projectile is given by

`R=(u^2\sin 2\theta )/(g)`

where u=launch velocity

`7.78=(u^2\sin (2* 32.9))/(9.8)`

`u^2=83.58`

`u=√(83.58)`

`u=9.142\ m/s`

(b)If u is increased by 8% then

new velocity is
`u'=9.87\ m/s`

`R'=(u'^2\sin 2\theta )/(g)`

`R'=9.07\ m`