A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The center of the cannon's target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon.
will the balloon hit the target? Can someone show me how to do it please?
Recall that the range of a missile is given by
R = (v^2 sin2θ)/g
plug in your numbers and see whether 41.0 <= R <= 43.0
so, what do you get for R?
I want to see 2019 physics passed question
To determine whether the water balloon will hit the target, we need to calculate its horizontal and vertical components of velocity as it leaves the water cannon. We can then use these components to determine the time it takes for the balloon to reach the ground and the horizontal distance it travels.
Step 1: Calculate the vertical component of velocity:
The vertical component of velocity is given by V_y = V * sin(θ), where V is the initial velocity of the balloon, and θ is the launch angle.
V_y = 34.0 m/s * sin(18°)
V_y ≈ 9.60 m/s
Step 2: Calculate the time of flight:
The time of flight is given by the equation t = 2 * V_y / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
t = 2 * 9.60 m/s / 9.8 m/s^2
t ≈ 1.96 s
Step 3: Calculate the horizontal component of velocity:
The horizontal component of velocity is given by V_x = V * cos(θ), where V is the initial velocity of the balloon, and θ is the launch angle.
V_x = 34.0 m/s * cos(18°)
V_x ≈ 31.51 m/s
Step 4: Calculate the horizontal distance traveled:
The horizontal distance traveled is given by d = V_x * t.
d = 31.51 m/s * 1.96 s
d ≈ 61.79 m
Since the center of the target is 42.0 m away, and the water balloon travels approximately 61.79 m horizontally, it will hit the target.
To determine whether the water balloon will hit the target or not, we need to analyze its projectile motion. Here are the steps to solve this problem:
Step 1: Break down the initial velocity into its horizontal and vertical components.
Given that the water balloon is fired at an angle of 18° above the ground with a velocity of 34.0 m/s, we can find the horizontal and vertical components using trigonometry.
Horizontal component (Vx): Vx = Velocity * cos(angle)
Vertical component (Vy): Vy = Velocity * sin(angle)
Step 2: Determine the time of flight.
The water balloon will experience the same time of flight in the horizontal direction as it does in the vertical direction. We can calculate the time of flight using the vertical component and the acceleration due to gravity.
Time of flight (t): t = (2 * Vy) / g
where g is the acceleration due to gravity (9.8 m/s²).
Step 3: Calculate the horizontal distance traveled by the water balloon.
The horizontal distance traveled by the water balloon can be found by multiplying the horizontal component of velocity by the time of flight.
Distance (d): d = Vx * t
Step 4: Check if the distance traveled exceeds the distance to the target.
Compare the calculated horizontal distance (d) with the given distance to the target (42.0 m).
If the distance traveled (d) is greater than or equal to the distance to the target (42.0 m), then the water balloon will hit the target. Otherwise, it will fall short.
By plugging in the given values and following these steps, you can determine whether the water balloon will hit the target.