The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.45 m/s at an angle of 53.0 degrees above the horizontal, and aims for a beetle on a leaf 4.00 cm above the water's surface.
At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time?
How much time will the beetle have to react?
I tried t=2.45sin(53)/9.81 and x=2.45cosine(53)*t but they both gave me wrong answers? What exactly do I need to do?
Wtf. Damon you are horrible at this. this is wrong again
Thx I got it!!!
The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.25m/s at an angle of 48.0 ∘ above the horizontal, and aims for a beetle on a leaf 3.50cm above the water's surface.
Well, it seems like you're fishing for the right answer but maybe using the wrong bait. Let's help you catch it.
To find the horizontal distance at which the archerfish should fire, we need to find the time it takes for the water stream to reach the beetle. We can use the equation:
h = (1/2)gt^2
where:
h = 4.00 cm = 0.04 m (height above the water's surface)
g = acceleration due to gravity = 9.81 m/s^2
t = time
Rearranging the equation, we get:
t = sqrt(2h/g)
Substituting the values, we find:
t = sqrt(2 * 0.04 / 9.81)
≈ 0.09 s
So, the beetle will have approximately 0.09 seconds to react.
Now, to find the horizontal distance, we can use the equation:
x = v * cos(θ) * t
where:
v = 2.45 m/s (speed of the water stream)
θ = 53.0 degrees (angle above the horizontal)
t = 0.09 s (time)
Substituting the values, we get:
x = 2.45 * cos(53.0) * 0.09
≈ 0.20 m
Therefore, the archerfish should fire at a horizontal distance of approximately 0.20 meters from the beetle to hit its target in the least time.
Remember, when calculating the time, make sure you're using the correct formula and converting units if necessary. And always aim for accuracy, even if you're using a water gun.
To determine the horizontal distance from the beetle that the archerfish should fire at and the time the beetle has to react, we can use the kinematic equations for projectile motion.
First, let's find the time it takes for the water stream to hit the beetle. We can use the equation:
y = (v₀y * t) + (0.5 * g * t²)
where:
- y is the vertical displacement, which is 4.00 cm or 0.04 m
- v₀y is the initial vertical velocity, which is the vertical component of the water's speed, v * sin(θ), where v is 2.45 m/s and θ is 53.0 degrees
- g is the acceleration due to gravity, approximately 9.81 m/s²
- t is the time we want to find
Rearranging the equation, we get:
0.04 = (2.45 * sin(53.0) * t) + (0.5 * 9.81 * t²)
Simplifying and moving all terms to one side, we have:
0.49 * t² + 1.226 * t - 0.04 = 0
Solving this quadratic equation using the quadratic formula, we find two possible values for t: t₁ ≈ 0.084 s and t₂ ≈ -2.884 s. Since time cannot be negative, we discard t₂.
Therefore, the beetle has approximately 0.084 seconds to react.
Next, let's find the horizontal distance from the beetle that the archerfish should fire at. We can use the equation:
x = v₀x * t
where:
- x is the horizontal distance we want to find
- v₀x is the initial horizontal velocity, which is the horizontal component of the water's speed, v * cos(θ), where v is 2.45 m/s and θ is 53.0 degrees
- t is the time we just found, approximately 0.084 s
Plugging in the values, we get:
x = (2.45 * cos(53.0)) * 0.084
Evaluating this expression, we find:
x ≈ 0.188 m
Therefore, the archerfish should fire at a horizontal distance of approximately 0.188 meters from the beetle in order to hit its target in the least time.
To hit in the least time get the target while the water is still rising, not coming back down,
h = .04 meters high
a = g = -9.8 m/s^2
vertical initial speed Vo = 2.45 sin 53 = 1.96 m/s
z = Zo + Vo t +.5 a t^2
.04 = 1.96 t - 4.9 t^2
4.9 t^2 - 1.96 t + .04 = 0
solve quadratic for t, use shortest of the two answers
then horizontal distance = t *2.45 cos 53