Question

In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 34.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 21.0 m. The first balloon launched by the Jackson team hits Walton dorm 1.80 s after launch, striking it halfway between the ground and the roof. Ignore air resistance. (a) Find the direction of the balloon's initial velocity. Give your answer as an angle (in degrees) measured above the horizontal. (b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon.

Answer 1

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

Step-by-step 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