A water balloon is launched at a speed of 23 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 20 m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.

Question

A water balloon is launched at a speed of 23 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 20 m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.

Answer 1

Horizontal problem:

u = 23 cos 36 until it stops

20 meters = u t
so
t = 20 /(23 cos 36)
use that t in the vertical problem

now vertical problem:

Vi = 23 sin 36 initial speed up

h = 0 + Vi t - 4.9 t^2

Answer 2

To solve this problem, we can break it down into horizontal and vertical components and use the equations of motion to find the height at which the water balloon hits the building.

First, let's find the time it takes for the water balloon to reach the building. We'll focus on the horizontal component first.

The horizontal component of the velocity (Vx) remains constant throughout the motion. We can find Vx using the initial velocity (23 m/s) and the angle (36 degrees) above the horizontal:

Vx = V * cos(θ)
Vx = 23 m/s * cos(36°)
Vx = 19.008 m/s (rounded to three significant figures)

Next, we can calculate the time it takes for the water balloon to reach the building:

Distance = Velocity * Time
20 m = 19.008 m/s * Time

Solving for Time:
Time = 20 m / 19.008 m/s
Time = 1.052 seconds (rounded to three significant figures)

Now, let's find the height at which the water balloon hits the building. We'll focus on the vertical component of the motion.

Using the vertical component of the velocity (Vy) and the time (1.052 seconds), we can determine the height:

First, we need to find Vy using the initial velocity (23 m/s) and the angle (36 degrees) above the horizontal:

Vy = V * sin(θ)
Vy = 23 m/s * sin(36°)
Vy = 13.589 m/s (rounded to three significant figures)

For vertical motion, we can use the following equation to find the height (h):

h = Vy * t + (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values:
h = 13.589 m/s * 1.052 s + (1/2) * 9.8 m/s^2 * (1.052 s)^2

Calculating:
h = 14.295 m (rounded to three significant figures)

Therefore, the water balloon will hit the building at a height of approximately 14.295 meters above the ground level.