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 ?

Answer 1

a)Vo = 9.58 m/s.

b)D= 9.04

The jump would be 1.26 m Longer.

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

Answer:

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

$R=\frac{u^2\sin 2\theta }{g}$

where u=launch velocity

$7.78=\frac{u^2\sin (2\times 32.9)}{9.8}$

$u^2=83.58$

$u=\sqrt{83.58}$

$u=9.142\ m/s$

(b)If u is increased by 8% then

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

$R'=\frac{u'^2\sin 2\theta }{g}$

$R'=9.07\ m$