A hawk is flying horizontally at 10 m/s in a straight line 200m above the ground. A mouse it was carrying is released from its grasp. The hawk continues on its path at the same speed for two seconds before attempting to retrieve its prey. To accomplis the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3m above the ground. Assuming no air resistance
a) find the diving speed of the hawk
b)what angle did the hawk make with the horizontal during its descent?
c)for how long did the mouse enjoy free fall?
a) Well, to find the diving speed of the hawk, we need to calculate its average velocity during those two seconds. Since the hawk is flying horizontally at a constant speed of 10 m/s, we can say its horizontal velocity remains the same. Now, let's focus on the vertical motion. From the given information, we know that the hawk descends 197 meters (200m - 3m) in 2 seconds. Since we're dealing with constant acceleration, we can use the equation:
d = v_i * t + (1/2) * a * t^2
where d is the displacement in the vertical direction, v_i is the initial velocity, t is time, and a is acceleration. Since the hawk starts from rest when it releases the mouse, the initial velocity is zero. Thus, our equation simplifies to:
d = (1/2) * a * t^2
Plugging in the known values, we get:
197m = (1/2) * a * (2s)^2
Now, we can solve for the acceleration, a:
a = (2 * 197m) / (2s)^2
a = 197m / 4s^2
a = 49.25m/s^2
Now that we have the acceleration, we can find the diving speed using the equation:
v_f = v_i + a * t
Since the hawk starts with a horizontal velocity of 10 m/s and continues flying horizontally throughout the descent, the initial vertical velocity is zero. Thus, our equation simplifies to:
v_f = a * t
v_f = (49.25m/s^2) * (2s)
v_f = 98.5m/s
So, the diving speed of the hawk is approximately 98.5 m/s.
b) To find the angle the hawk made with the horizontal during its descent, we can use tangent:
tan(θ) = vertical speed / horizontal speed
tan(θ) = 98.5m/s / 10m/s
tan(θ) = 9.85
Now, we can take the inverse tangent to find the angle θ:
θ ≈ tan^(-1)(9.85)
θ ≈ 84.24 degrees
So, the hawk made an angle of approximately 84.24 degrees with the horizontal during its descent.
c) The time the mouse enjoyed free fall is the time it took for the hawk to retrieve it, which is 2 seconds. That's two whole seconds of feeling weightless and free-falling, just like a skydiver without a parachute! I hope those were two memorable seconds for the mouse.
To solve this problem, we can use the equations of motion to calculate the diving speed of the hawk and the time it took for the mouse to free fall.
a) To find the diving speed of the hawk, we can use the equation of motion for vertical displacement:
s = ut + (1/2)gt^2
where s is the vertical displacement, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
The initial vertical velocity of the hawk is 0, as it was flying horizontally. The vertical displacement is -3m (since it descended 3m to recapture the mouse), and the time is 2 seconds (as given in the problem). The acceleration due to gravity is approximately 9.8 m/s^2.
Plugging in the values, we get:
-3 = 0 + (1/2)(9.8)(2)^2
-3 = 0 + 9.8(4)
-3 = 39.2
So, the diving speed of the hawk is 39.2 m/s.
b) To find the angle the hawk makes with the horizontal during its descent, we can use the trigonometric relationship between the vertical and horizontal components of the velocity.
tan(theta) = v_vertical / v_horizontal
Here, we know the diving speed of the hawk is 39.2 m/s and the horizontal speed of the hawk is 10 m/s.
tan(theta) = (39.2 m/s) / (10 m/s)
tan(theta) = 3.92
To find the angle theta, we can take the inverse tangent (arctan) of both sides:
theta = arctan(3.92)
Using a scientific calculator or an online tool, we find:
theta ≈ 76.3 degrees
So, the hawk makes an angle of approximately 76.3 degrees with the horizontal during its descent.
c) To find the time the mouse enjoyed free fall, we can use the equation of motion for vertical displacement:
s = ut + (1/2)gt^2
where s is the vertical displacement, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
Before being released, the mouse was being carried by the hawk. So, the initial vertical velocity of the mouse is the same as the hawk's horizontal velocity, which is 10 m/s. The vertical displacement is -200m (as it was initially 200m above the ground and descended 200m to reach the ground). We need to find the time the mouse took to fall this distance.
Plugging in the values, we get:
-200 = 10t + (1/2)(9.8)t^2
-200 = 10t + 4.9t^2
Rearranging the equation and setting it equal to zero:
4.9t^2 + 10t + 200 = 0
This is a quadratic equation, which can be solved using the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / (2a)
Here, a = 4.9, b = 10, and c = 200.
t = [-10 ± √(10^2 - 4(4.9)(200))] / (2(4.9))
t = (-10 ± √(100 - 3920)) / 9.8
t = (-10 ± √(-3820)) / 9.8
Since the square root of a negative number is not defined in the real number system, it means that the quadratic equation has no real solutions. Thus, the equation represents a situation where the mouse never reaches the ground, and it does not enjoy free fall.
Therefore, the mouse did not enjoy any free fall time.
a) To find the diving speed of the hawk, we can use the concept of conservation of energy. Since there is no air resistance, the total mechanical energy of the system (hawk + mouse) is conserved.
Initially, the hawk has only kinetic energy, so we can write the equation as:
1/2 * m_hawk * v_initial^2 = m_hawk * g * h_initial
Where:
m_hawk is the mass of the hawk (which is assumed to be equal to the mass of the mouse).
v_initial is the initial speed of the hawk (which is given as 10 m/s).
g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
h_initial is the initial height of the hawk above the ground (which is given as 200 m).
Simplifying the equation:
1/2 * v_initial^2 = g * h_initial
Plugging in the given values:
1/2 * 10^2 = 9.8 * 200
Solving for v_initial:
v_initial = sqrt((9.8 * 200) / 0.5)
v_initial ≈ 19.8 m/s
Therefore, the diving speed of the hawk is approximately 19.8 m/s.
b) To find the angle the hawk makes with the horizontal during its descent, we can use trigonometry. The diving speed can be split into its horizontal and vertical components.
The horizontal component of the diving speed remains constant at 10 m/s. The vertical component is given by the equation:
v_vertical = v_diving * sin(theta)
Where:
v_diving is the speed of the hawk during its descent (which we need to find).
theta is the angle the hawk makes with the horizontal during its descent.
Since the hawk descends for 2 seconds and catches the mouse 3 m above the ground, we can find the vertical component of the diving speed using the equation:
h_descend = v_vertical * t + (1/2) * g * t^2
Plugging in the values:
3 = v_diving * sin(theta) * 2 + (1/2) * 9.8 * 2^2
Simplifying the equation:
3 = 2 * v_diving * sin(theta) + 19.6
Since v_diving = 19.8 m/s (as calculated in part a), we can substitute it into the equation to solve for sin(theta):
3 = 2 * 19.8 * sin(theta) + 19.6
sin(theta) = (3 - 19.6) / (2 * 19.8)
Simplifying further:
sin(theta) ≈ -0.558
To find the angle theta, we can take the inverse sin of -0.558:
theta ≈ sin^(-1)(-0.558)
theta ≈ -33.9°
Therefore, the angle the hawk makes with the horizontal during its descent is approximately -33.9° (measured below the horizontal).
c) The mouse would enjoy free fall until it is caught by the hawk. From the given information, we know that the hawk descends for 2 seconds before recapturing the mouse. Therefore, the mouse would experience free fall for 2 seconds.
OK, so first I'd do the equation in the Y direction to find the answer to part 3).
X = Xo + Vot + 1/2at^2
3 = 180 + (0*t) + (1/2 * -9.8 * t^2)
3 = 180 - 4.9t^2
4.9 t^2 = 177
t^2 = 36.12
t = 6.01s
Now we go back to part 1). We know the Hawk waited 2s before starting its dive so it dove for 6.01 - 2 = 4.01 s. We also know that during that time, it traveled 180 - 3 = 177 m vertically. But, since there's no air resistance, the mouse kept traveling horizontally too! And, during those 6.01s, it traveled 10 * 6.01 = 60.1 m forward. Since the Hawk traveled 20 m in the 2 s it waited, it must have traveled 60.1 - 20 = 40.1 m horizontally during it's dive!
OK, so it dove 177 m vertically while traveling 40.1 m horizontally. Pythagorean theorem then says d = sqrt(177^2 + 40.1^2) = 181.5 m. Since it did that in 4.01 s, its speed must have been 181.5 / 4.01 = 45.26 m/s.
We can calculate the angle... Draw a picture and you may see...
tan(theta) = 177 / 40.1 = 4.414
theta = 1.348 rad or 77.2°