Question

Una atleta de salto a lo largo salta a una velocidad de 28 ft/s con un angulo inicial de 40 grados. Determine la medida del salto horizontal que logro en pies.

A long jump athlete jumps at a speed of 28 ft / s with an initial angle of 40 degrees. Determine the measure of the horizontal jump you achieved in feet.

Answer 1

Given a long jump athlete jumps at a speed of 28ft/s with initial angle of 40^o.

Intial velocity U=28 ft/s

Initial angle =40^o

Horizontal distance achieved by athlete (Range) =-----?

This problem is related to projectile in a plane

At the initial point

velocity in X direction =UCos

velocity in Y direction =USin

Using the relation S=ut+(1/2)at²

In X direction

u=UCos a=0

Displacement in X direction S_x=UCost --------------------------------(1)

Displacement in Y direction S_y=USint-(1/2)at²-----------------------(2)

To get the complete time of flight use equation no 2

AT the final point

S_y=0

t=time of flight (T)

a=g

substitute these in eq 2 we get

0=USinT-(1/2)gT²

solving for T we get

T=(2USin)/g

The distance achieved (also called "Range") by the athlete in the long jump can be measured by Using the equation 1 and total time T

S_x=UCost

S_x=Measure of horizontal jump

U=Initial velocity =28 ft/s

=Initial angle= 40^o

^{T= total time for the jump =(2USin)/g}

g= 32.174 ft/sec²

^{substituting all the above we get}

^{S_x=UCos *(2USin)}/g

S_x=(U²2SinCos)/g

S_x=(U²Sin2)/g

= (28²Sin(2*40^o))/32.174

=23.99 ft ~24 feets

Therefore the measure of the horizontal jump achieved by the athlete is 24 feet (approx)