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,

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

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

Velocity = 90.0m/s

Angle =$23.0°$

Distance = 114 m

For the vertical motion velocity component is $v\sin \mathrm{θ}$and for the horizontal motion the velocity component is $v\cos \mathrm{θ}$.

For the vertical motion from the second law of motion,

$\mathrm{s}=\mathrm{\pm ¹²õ¾\pm ²Ôθ³Ù}+\frac{1}{2}{\mathrm{at}}^2$

Substitute 114m for s, 90m/s for v, $23°$for $\mathrm{θ}$and $9.8\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$$\begin{gathered} 114=90×\mathrm{t}+\frac{1}{2}×9.8×{\mathrm{t}}^2 \\ \mathrm{t}=9.60\mathrm{s} \end{gathered}$$

In the time 9.60s the horizontal distance traveled by suitcase,

$\mathrm{s}=\mathrm{\pm ¹³¦´Ç²õθ³Ù}+\frac{1}{2}{\mathrm{at}}^2$

Substitute 90.0m/s for v, $23°$for $\mathrm{θ}$, 9.60s for t and $0m/s^2$ for a in the above equation.

$$\begin{gathered} \mathrm{s}=90.0×\cos 23°×9.60+\frac{1}{2}×0×{\left(9.60\right)}^2 \\ \mathrm{s}=795\mathrm{m} \end{gathered}$$

Therefore the suitcase will land 795m far from the dog.