Question

An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?

Answer 1

Given data:

* The initial velocity of the jumper is u = 9.1 m/s.

* The horizontal range in the given case is 8 m.

Solution:

(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,

`R=ut+(1)/(2)at^2`

where a is the acceleration of the jumper in the horizontal direction,

As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.

Substituting the known values,

`\begin{gathered} 8=9.1* t \\ t=(8)/(9.1) \\ t=0.88\text{ s} \end{gathered}`

Thus, the time for which the jumper remains in the air is 0.88 seconds.

(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,

`v_y-u_y=gt^(\prime)_{}\ldots\ldots\ldots(1)`

where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,

The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.

The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.

As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,

`\begin{gathered} t^(\prime)=(t)/(2) \\ t^(\prime)=(0.88)/(2) \\ t^(\prime)=0.44\text{ s} \end{gathered}`

Substituting the known values in the equation (1),

`\begin{gathered} 0-u_y=-9.8*0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}`

By the kinematics equation, the maximum height reached by the jumper is,

`h=u_yt^(\prime)+(1)/(2)gt^(\prime)^2`

Substituting the known values,

`\begin{gathered} h=4.312*0.44+(1)/(2)*(-9.8)*(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}`

Thus, the maximum height reached by the jumper is 0.95 meters.