a) Horizontal distance covered: 8.0 m
b) The final speed is 14.3 m/s
Explanation:
a)
The motion of the package is a projectile motion, so it consists of two independent motions: Â
- A uniform motion (constant velocity) along the horizontal direction Â
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction Â
First of all, we have to find the time of flight of the package. This is given by the vertical motion, and we can do it by using the suvat equation:
[tex]s=u_y t+\frac{1}{2}at^2[/tex]
where:
s = 8.65 m is the vertical displacement of the package (the initial height from which is thrown)
[tex]u_y=0[/tex] is the initial vertical velocity (the package is thrown horizontally)
t is the time of flight
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for the time of flight we find:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(8.65)}{9.8}}=1.33 s[/tex]
Therefore the package lands after 1.33 s.
Now we can study the horizontal motion: the package moves horizontally with a constant velocity of
[tex]v=21.5 km/h \cdot \frac{1000}{3600}=6.0 m/s[/tex]
And therefore the distance it covers in 1.33 s is
[tex]d=v_x t =(6.0)(1.33)=8.0 m[/tex]
So the package covers a horizontal distance of 8.0 m.
b)
As we said previously, the horizontal velocity of the package is constant, and it is
[tex]v_x = 6.0 m/s[/tex]
On the other hand, the vertical velocity is constantly changing, and it is given by
[tex]v_y = u_y + at[/tex]
where
[tex]u_y = 0[/tex] (initial vertical velocity)
[tex]a=g=9.8 m/s^2[/tex]
Substituting t = 1.33 s, we find the final vertical velocity:
[tex]v_y = 0 +(9.8)(1.33)=13.0 m/s[/tex] (in the downward direction)
Therefore, the final speed is given by:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{6.0^2+13.0^2}=14.3 m/s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
Answer:
(a) [tex]x=7.93\ m[/tex]
(b) [tex]v=15.25\ m/s[/tex]
Explanation:
Horizontal Launching
When an object is released from certain height h with a (only) horizontal speed vo, it describes a curved trajectory with a constant speed in the horizontal direction and a variable speed in the vertical direction, constantly changed by the acceleration of gravity. Their equations are
[tex]v_x=v_o[/tex]
[tex]v_y=g.t[/tex]
The distances traveled in both directions are
[tex]x=v_o.t[/tex]
[tex]\displaystyle y=h-\frac{gt^2}{2}[/tex]
(a)
The package dropped by the drone will land when it travels down all its initial height, thus the flight time must comply:
[tex]\displaystyle 0=h-\frac{gt_f^2}{2}[/tex]
Solving for [tex]t_f[/tex] :
[tex]\displaystyle t_f=\sqrt{\frac{2h}{g}}[/tex]
The data is: h=8.65 m, v_o=21.5 Km/h = 5.972 m/s
[tex]\displaystyle t_f=\sqrt{\frac{2(8.65)}{9.8}}=1.33\ sec[/tex]
Thus, the horizontal distance is
[tex]x=(5.972)*(1.33)=7.93\ m[/tex]
[tex]\boxed{x=7.93\ m}[/tex]
(b)
The horizontal speed at landing time is (constant)
[tex]v_x=5.972\ m/s[/tex]
The vertical speed is
[tex]v_y=(9.8)*(1.33)=13.024\ m/s[/tex]
The magnitude of the total speed is
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{(7.93)^2+(13.024)^2}=15.25\ m/s[/tex]
[tex]\boxed{v=15.25\ m/s}[/tex]
