Answer:
35.489km
Explanation:
a diagram illustrating the question is attached
now, applying cosine rule to find the displacement c
cosine rule;
c²=a²+b²-2abCos ∅
c²=50²+80²-2(50)(80)Cos 38°
c²=2500 + 6400 - (8000×0.9951)
c²=8900-7640.589
c²=1259.411
c=
c=35.49km
Answer:
a) The direction of the initial velocity of the first balloon is 18.5º
b) The initial speed of the second balloon is 19.1 m/s
Explanation:
The equations for the vector position in a parabolic motion are as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = vector position
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity
a) At t = 1.80 s the vector r will be as shown in the figure (in yellow). The x-component of the vector r, rx, is 34.0 m and the y-component, ry, is -4.5 m. The reference system is located at the edge of the roof of the Jackson building.
Then, at 1.80 s, r will be:
r = (34.0 m, - 4.50 m)
Using the equations for each component we can obtain v0 and α:
Using the x- component:
x = x0 + v0 · t · cos α
34.0 m = v0 · 1.80 s · cos α (x0 = 0 because the origin of the reference system is located at the launching point)
34.0 m / (1.80 s · cos α) = v0
Replacing v0 in the equation of the y-component:
y = y0 + v0 · t · sin α + 1/2 · g · t²
-4.50 m = (34.0 m /cos α) · sin α + 1/2 · - (9.80 m/s²) · (1.80 s)²
(sin α / cos α = tan α)
-4.50 m = 34.0 m · tan α - 1/2 · 9.80m/s² · (1.80 s)²
solving for tan α
tan α = 0.334
α = 18.5º
The direction of the balloon´s initial velocity is 18.5º above the horizontal.
b) In green is the trajectory of the second balloon. r2 is the final vector of the second balloon and its components in x and y are 34 m and - 6 m (21 m- 15 m) respectively (see figure).
Then:
r2 = (34.0 m, - 6.00 m) Now the reference system is located at the launching point of the second balloon.
We can proceed in the same way as in the point a) only that now we have the angle but do not have the time nor the initial velocity.
Using the equation of the x-component of vector r2:
x = x0 + v0 · t · cos α
34.0 m = v0 · t · cos 18.5º
34.0 m /(t · cos 18.5º) = v0
Replacing v0 in the equation of the y-component:
y = y0 + v0 · t · sin α + 1/2 · g · t²
-6.00 m = 34.0 m · tan 18.5º - 1/2 · 9.8 m/s² · t²
-2 (-6.00 m - 34 m · tan 18.5º) / 9.8 m/s² = t²
t = 1.88 s
Then, the speed will be:
v0 = 34.0 m /(1.88 s · cos 18.5º) = 19.1 m/s
