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? justify your response with calculation.
my answer would be:
horizontal component: v1x = 34 cos18=32.3m/s
vertical component: v2x 34 sin18 = 10.5m/s
v2 = v1+a(▲t)
0=10.5+(-9.8)(▲t)
▲t=10.5/9.8
▲t=1.07s
time taken to reach max height = 1.07s
time of flight = 2 x 1.07 = 2.14s
horizontal distance
▲d = v1x *▲t
▲d = (32.3)(2.14)
▲d=69.1
∴ It will miss the target which is 42 m away from the cannon.
Range = Vo^2*sin(2A)/g = 34^2*sin(36)/9.8 = 69.3 m.
Yes, you are correct.
Well, it seems like the water balloon missed the target by quite a splash! The poor little balloon was fired with a speed of 34.0 m/s at an angle of 18° above the ground. Let's break it down.
First, we need to find the horizontal and vertical components of the initial velocity. The horizontal component (v1x) is determined by multiplying the speed (34.0 m/s) by the cosine of the angle (18°), which gives us a value of 32.3 m/s. As for the vertical component (v1y), it is found by multiplying the speed (34.0 m/s) by the sine of the angle (18°), yielding a value of 10.5 m/s.
Now, let's calculate the time it takes for the balloon to reach its maximum height. We can use the equation v2 = v1 + a(Δt), where v2 is the final vertical velocity (0 m/s), v1 is the initial vertical velocity (10.5 m/s), and a is the acceleration due to gravity (-9.8 m/s^2). Solving for Δt, we find that it takes approximately 1.07 seconds to reach the maximum height.
Since the balloon's flight path is symmetric, the total time of flight is twice the time taken to reach the maximum height, which results in a flight time of 2.14 seconds.
Now, let's calculate the horizontal distance traveled by the water balloon using the equation Δd = v1x * Δt, where Δd is the horizontal distance, v1x is the horizontal component of the initial velocity (32.3 m/s), and Δt is the total time of flight (2.14 seconds). Multiplying these values together, we find that the balloon traveled approximately 69.1 meters horizontally.
So, in conclusion, the poor water balloon missed the target by quite a distance. The target, which was painted on the asphalt 42.0 meters away from the water cannon, was just too elusive for the slippery little balloon. Better luck next time, balloon!
Your calculations are correct. The horizontal distance traveled by the water balloon is 69.1 meters, which is greater than the distance to the target (42 meters). Therefore, the water balloon will miss the target.
To determine whether the water balloon will hit the target or not, we need to calculate the horizontal distance the balloon will travel. Here's how to do it:
First, we need to calculate the horizontal and vertical components of the initial velocity of the water balloon.
The horizontal component can be found using the equation: v1x = v * cos(θ), where v is the magnitude of the initial velocity and θ is the angle above the ground. In this case, v = 34.0 m/s and θ = 18°.
v1x = 34.0 * cos(18°) = 32.3 m/s.
The vertical component can be found using the equation: v2y = v * sin(θ).
v2y = 34.0 * sin(18°) = 10.5 m/s.
Next, we can calculate the time taken for the balloon to reach its highest point by using the vertical component of velocity and the acceleration due to gravity.
v2y = v1y + a * t,
where a is the acceleration due to gravity (-9.8 m/s^2) and t is the time taken.
0 = 10.5 - 9.8 * t.
9.8 * t = 10.5.
t ≈ 1.07 s.
Since the projectile reaches its highest point at half of the total time of flight, we can determine the total time of flight:
time of flight = 2 * t ≈ 2 * 1.07 s = 2.14 s.
Finally, we can calculate the horizontal distance traveled by the water balloon using the horizontal component of velocity and the time of flight:
▲d = v1x * t,
▲d = 32.3 m/s * 2.14 s,
▲d ≈ 69.1 m.
The target is painted on the asphalt 42.0 m away from the water cannon, and since the horizontal distance traveled by the water balloon is greater than the distance to the target, which is 69.1 m > 42.0 m, the balloon will miss the target.