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,

$H=\frac{u^2\sin (2\theta)}{g}$

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

Substituting the known values,

$\begin{gathered} H=\frac{120^2\times\sin (2\times30^{\circ})}{9.8} \\ H=1272.53\text{ m} \end{gathered}$

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