Question

A long jumper with a speed of 120m/s at an angle of 30 degrees with respect to the horizontal. how far did he land in the sand from his jump spot?

Answer 1

Given data:

* The initial velocity of the jumper is,

`u=120ms^(-1)`

* The angle between the initial velocity and the horizontal line is,

`\theta=30^(\circ)`

Solution:

The horizontal range of the projectile motion by the jumper is,

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

where g is the accleration due to gravity, and H is the horizontal range.

Substituting the known values,

`\begin{gathered} H=(120^2*\sin (2*30^(\circ)))/(9.8) \\ H=1272.53\text{ m} \end{gathered}`

Thus, the distance of the jump from its spot is 1272.53 m.