When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of 13.3 m/s, at an angle of 34.0° above the horizontal. (a) What is the range of his leap and (b) for how much time is he in the air?
a. Range = Vo^2*sin(2A)/g = 13.3^2*sin(68)/9.8 = 16.7 m.
b. Xo = 13.3*Cos34 = 11.0 m/s. = Hor. component of Vo.
Range = Xo*T = 16.7,
11.0 * T = 16.7,
T = 1.52 s.
To find the range and time of flight, we need to split the initial velocity into horizontal and vertical components.
The horizontal component (Vx) can be found using the equation Vx = V * cos(θ):
Vx = 13.3 m/s * cos(34.0°) ≈ 11.0 m/s
The vertical component (Vy) can be found using the equation Vy = V * sin(θ):
Vy = 13.3 m/s * sin(34.0°) ≈ 7.1 m/s
(a) To find the range, we use the equation R = Vx * t, where t is the time of flight.
Let's consider the vertical motion, where the initial vertical velocity (Vy) is 7.1 m/s, the final vertical velocity (Vfy) is -7.1 m/s, and the acceleration (a) is -9.8 m/s² (due to gravity). We can use the equation Vfy = Vy + at to find the time of flight (t).
Vfy = 0 m/s (at the top of the flight)
0 = 7.1 m/s + (-9.8 m/s²) * t
-7.1 m/s = -9.8 m/s² * t
t = -7.1 m/s / (-9.8 m/s²) ≈ 0.724 s (upward motion)
The total time of flight (T) will be twice the time of upward motion, since the hare will be in the air both while jumping up and falling back down.
T = 2 * t = 2 * 0.724 s ≈ 1.448 s (total time of flight)
Now, let's find the range (R) using the horizontal velocity (Vx) and the total time of flight (T).
R = Vx * T
R = 11.0 m/s * 1.448 s ≈ 15.92 m
The range of the leap is approximately 15.92 meters.
(b) The time of flight is approximately 1.448 seconds.
To find the range of the greyhound's leap, we need to determine the horizontal distance covered during the leap. The range is the total horizontal distance traveled by the object.
To calculate the range, we can use the horizontal component of the initial velocity (Vx) and the time of flight (t).
(a) Range of the leap:
The horizontal component of the initial velocity can be found using trigonometric functions. Since the greyhound leaped at an angle of 34.0° above the horizontal, we can calculate the horizontal component as:
Vx = V * cos(theta)
where V is the speed of 13.3 m/s and theta is the angle of 34.0°.
Vx = 13.3 m/s * cos(34.0°)
Vx ≈ 10.99 m/s
Now, we can calculate the range using the formula:
Range = Vx * t
Substituting the value of Vx, we have:
Range = 10.99 m/s * t
(b) Time in the air:
To calculate the time of flight, we can use the vertical component of the initial velocity (Vy) and the acceleration due to gravity (g = 9.8 m/s²). The time of flight can be determined using the formula:
Vy = V * sin(theta)
where V is the speed of 13.3 m/s and theta is the angle of 34.0°.
Vy = 13.3 m/s * sin(34.0°)
Vy ≈ 7.25 m/s
Next, we can determine the time of flight using the formula:
t = 2 * Vy / g
Substituting the values, we have:
t = 2 * 7.25 m/s / 9.8 m/s²
t ≈ 1.48 s
Therefore, the answers to the given questions are:
(a) The range of the leap is approximately 10.99 meters.
(b) The greyhound is in the air for approximately 1.48 seconds.