A water balloon is launched at a speed of 42 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 27 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.
u = 42 cos 36 horizontal speed
so
t = 27/(42 cos 36)
Vi = 42 sin 36 initial speed up
h = Vi t - 4.9 t^2
To find the height above the ground level where the water balloon hits the building, we can use the equations of projectile motion. The vertical motion of the water balloon can be analyzed separately from the horizontal motion.
First, let's find the time it takes for the water balloon to reach the building. We'll use the horizontal motion equation:
d = v * t
where:
d is the horizontal distance (27 m)
v is the initial horizontal velocity of the water balloon (42 m/s)
t is the time we want to find
Plugging in the known values:
27 = 42 * t
Solving for t:
t = 27 / 42
t ≈ 0.643 seconds (rounded to three decimal places)
Now, let's find the vertical distance the water balloon travels in that time. We'll use the vertical motion equation:
y = v * t - (1/2) * g * t^2
where:
y is the vertical distance
v is the initial vertical velocity of the water balloon (42 * sin(36°))
t is the time (0.643 seconds)
g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the known values:
y = (42 * sin(36°)) * 0.643 - (1/2) * 9.8 * (0.643)^2
Calculating:
y ≈ 14.876 - 2.536
y ≈ 12.34 meters (rounded to three decimal places)
Therefore, the water balloon hits the building at a height of approximately 12.34 meters above the ground level.
To solve this problem, we can break it down into two components: horizontal and vertical motion. We'll start by finding the time it takes for the water balloon to reach the building, and then use that time to find the height at which it hits the building.
First, let's find the time of flight (t) for the water balloon. We can use the horizontal component of its initial velocity (42 m/s) and the distance it travels horizontally (27 m) to calculate this.
The horizontal motion is not affected by gravity, so we can use the following equation:
distance = speed * time
Rearranging the equation to solve for time:
time = distance / speed
Plugging in the values:
time = 27 m / 42 m/s ≈ 0.643 s
Now that we have the time of flight, we can find the height at which the water balloon hits the building. We need to consider the vertical motion of the water balloon.
The vertical motion is affected by gravity, so we can use the following equations of motion:
vertical displacement = initial vertical velocity * time + (1/2) * acceleration due to gravity * time^2
The initial vertical velocity is the vertical component of the initial velocity, which can be calculated using trigonometry:
initial vertical velocity = initial velocity * sin(angle)
Plugging in the values:
initial vertical velocity = 42 m/s * sin(36°) ≈ 24.087 m/s
Now we can substitute the values into the equation:
vertical displacement = (24.087 m/s) * (0.643 s) + (1/2) * (-9.8 m/s^2) * (0.643 s)^2
Simplifying:
vertical displacement ≈ 15.512 m
Therefore, the water balloon hits the building at a height of approximately 15.512 meters above the ground level. Rounded to three significant figures, the height is 15.5 meters.