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
[tex]v_x = u cos \theta[/tex]
where
u = 8.3 m/s is the initial velocity of the balloon
[tex]\theta=27^{\circ}[/tex] is the angle of projection
Substituting,
[tex]v_x = (8.3)(cos 27^{\circ})=7.4 m/s[/tex]
b)
The vertical component of the initial velocity of a projectile is given by
[tex]u_y = u sin \theta[/tex]
where
u is the initial velocity
[tex]\theta[/tex] is the angle of projection
Here we have
u = 8.3 m/s
[tex]\theta=27^{\circ}[/tex]
Substituting,
[tex]u_y = (8.3)(sin 27^{\circ})=3.8 m/s[/tex]
c)
The vertical component of the velocity of the balloon follows the suvat equation
[tex]v_y = u_y - gt[/tex]
where
[tex]v_y[/tex] is the vertical velocity at time t
[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The balloon reaches the maximum height when the vertical velocity becomes zero:
[tex]v_y = 0[/tex]
So we get:
[tex]0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s[/tex]
d)
The maximum height of the balloon can be calculated using the suvat equation:
[tex]s=u_y t - \frac{1}{2}gt^2[/tex]
where
[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
t = 0.39 s is the time at which the highest point is reached
Substituting,
[tex]s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m[/tex]
e)
The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by
[tex]t=\frac{2u_y}{g}[/tex]
where
[tex]u_y[/tex] is the initial vertical velocity
[tex]g[/tex] is the acceleration of gravity
Here we have
[tex]u_y = 3.8 m/s[/tex]
[tex]g=9.8 m/s^2[/tex]
Substituting,
[tex]t=\frac{2(3.8)}{9.8}=0.78 s[/tex]
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:
[tex]d=v_x t[/tex]
where
[tex]v_x[/tex] is the horizontal velocity
t is the time of flight
Here we have
[tex]v_x = 7.4 m/s[/tex]
[tex]t = 0.78 s[/tex]
Substituting,
[tex]d=(7.4)(0.78)=5.77 m[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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