91Ó°ÊÓ

If a particle is thrown in projectile motion, then the velocity of the particle will contain two components, first one is horizontal and other one is vertical.

According to the Newton’s laws of motion,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Heres is the displacement,u is the initial velocity,a is the acceleration andt is the time.

Distance of the helicopter from the ground =30.0 m

Vertical component of the velocity${\mathrm{v}}_{\mathrm{y}}=10.0\mathrm{m}/\mathrm{s}$

Horizontal component of the velocity${\mathrm{v}}_{\mathrm{x}}=15.0\mathrm{m}/\mathrm{s}$

(a)

For the vertical motion from the law of motion of Newton,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Substitute 30.0m for s, 10 m/s foru and$9.8m/s^2$ fora in the above equation.

$$\begin{gathered} 30.0=10×\mathrm{t}+\frac{1}{2}×9.8×{\mathrm{t}}^2 \\ \mathrm{t}=3.70\mathrm{s} \end{gathered}$$

In the time 3.70s the horizontal distance travelled by her,

$s=ut+\frac{1}{2}a{t}^2$

Substitute 15.0m/s for u, 3.70s fort and$9.8m/s^2$ forg in the above equation.

$$\begin{gathered} \mathrm{s}=15.0×3.70+\frac{1}{2}×9.8×{\left(3.70\mathrm{s}\right)}^2 \\ \mathrm{s}=55.5\mathrm{m} \end{gathered}$$

Therefore, the mats should be placed at a distance 55.5m from the south.

(b)

x-t Graph

y-t Graph

$y_x-t$Graph

$v_y-t$Graph