a) The horizontal component of the velocity is 7.4 m/s
b) The vertical component of the velocity is 3.8 m/s
c) The balloon reaches the highest point after 0.39 s
d) The maximum height is 0.74 m
e) The total time of flight is 0.78 s
f) The range of the balloon is 5.77 m
Explanation:
a)
The motion of the balloon is the motion of a projectile, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
The horizontal component of the velocity (which is constant) is given by
![v_x = u cos \theta](https://tex.z-dn.net/?f=v_x%20%3D%20u%20cos%20%5Ctheta)
where
u = 8.3 m/s is the initial velocity of the balloon
is the angle of projection
Substituting,
![v_x = (8.3)(cos 27^{\circ})=7.4 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20%288.3%29%28cos%2027%5E%7B%5Ccirc%7D%29%3D7.4%20m%2Fs)
b)
The vertical component of the initial velocity of a projectile is given by
![u_y = u sin \theta](https://tex.z-dn.net/?f=u_y%20%3D%20u%20sin%20%5Ctheta)
where
u is the initial velocity
is the angle of projection
Here we have
u = 8.3 m/s
![\theta=27^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D27%5E%7B%5Ccirc%7D)
Substituting,
![u_y = (8.3)(sin 27^{\circ})=3.8 m/s](https://tex.z-dn.net/?f=u_y%20%3D%20%288.3%29%28sin%2027%5E%7B%5Ccirc%7D%29%3D3.8%20m%2Fs)
c)
The vertical component of the velocity of the balloon follows the suvat equation
![v_y = u_y - gt](https://tex.z-dn.net/?f=v_y%20%3D%20u_y%20-%20gt)
where
is the vertical velocity at time t
is the initial vertical velocity
is the acceleration of gravity
The balloon reaches the maximum height when the vertical velocity becomes zero:
![v_y = 0](https://tex.z-dn.net/?f=v_y%20%3D%200)
So we get:
![0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s](https://tex.z-dn.net/?f=0%3Du_y%20-gt%5C%5Ct%3D%5Cfrac%7Bu_y%7D%7Bg%7D%3D%5Cfrac%7B3.8%7D%7B9.8%7D%3D0.39%20s)
d)
The maximum height of the balloon can be calculated using the suvat equation:
![s=u_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3Du_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
is the initial vertical velocity
is the acceleration of gravity
t = 0.39 s is the time at which the highest point is reached
Substituting,
![s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m](https://tex.z-dn.net/?f=s%3D%283.8%29%280.39%29-%5Cfrac%7B1%7D%7B2%7D%289.8%29%280.39%29%5E2%3D0.74%20m)
e)
The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by
![t=\frac{2u_y}{g}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2u_y%7D%7Bg%7D)
where
is the initial vertical velocity
is the acceleration of gravity
Here we have
![u_y = 3.8 m/s](https://tex.z-dn.net/?f=u_y%20%3D%203.8%20m%2Fs)
![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
Substituting,
![t=\frac{2(3.8)}{9.8}=0.78 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2%283.8%29%7D%7B9.8%7D%3D0.78%20s)
f)
The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:
![d=v_x t](https://tex.z-dn.net/?f=d%3Dv_x%20t)
where
is the horizontal velocity
t is the time of flight
Here we have
![v_x = 7.4 m/s](https://tex.z-dn.net/?f=v_x%20%3D%207.4%20m%2Fs)
![t = 0.78 s](https://tex.z-dn.net/?f=t%20%3D%200.78%20s)
Substituting,
![d=(7.4)(0.78)=5.77 m](https://tex.z-dn.net/?f=d%3D%287.4%29%280.78%29%3D5.77%20m)
Learn more about projectile motion:
brainly.com/question/8751410
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