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 ?

Explanation:

Given

Inclination

Distance of landing point

Considering athlete to be an Projectile

range of projectile is given by

where u=launch velocity

(b)If u is increased by 8% then

new velocity is

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.

d. air pressure decreases as altitude increases.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j