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Answer:
Total distance traveled by bird is 10.5 km.
Explanation:
Given that,
Velocity of runner= 8.4 km/hr
Distance = 6.3 km
Velocity of bird = 42 km/hr
Suppose, How far does the bird travel?
We need to calculate the time of bird
Using formula of time
Put the value into the formula
We need to calculate the distance runner travels during time period
Using formula of distance
We need to calculate the distance of runner to FL
Using distance
We need to calculate the net closing speed between runner and bird
Using speed
We need to calculate the time for runner and bird to meet
Using formula of time
Put the value into the formula
We need to calculate the distance covered by runner
Using formula of distance
Put the value into the formula
We need to calculate the distance covered by bird to runner
Using formula of distance
Put the value into the formula
We need to calculate the total distance
Using distance
Hence, Total distance traveled by bird is 10.5 km.
Answer:
The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.
Explanation:
Hi there!
The position vector of the ball is described by the following equation:
r = (x0 + v0x · t + 1/2 · ax · t², y0 + v0y · t + 1/2 · ay · t², z0 + v0z · t + 1/2 · g · t²)
Where:
r = poisition vector of the ball at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity (eastward).
t = time.
ax = horizontal acceleration (eastward).
y0 = initial horizontal position.
v0y = initial horizontal velocity (southward).
ay = horizontal acceleration (southward)
z0 = initial vertical position.
v0z = initial vertical velocity.
g = acceleration due to gravity.
We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.
Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.
y = z0 + v0z · t + 1/2 · g · t² (z0 = 0)
0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²
0 = t (48 ft/s - 16 ft / s² · t) (t= 0, the origin point)
0 = 48 ft/s - 16 ft / s² · t
- 48 ft/s / -16 f/s² = t
t = 3.0 s
Now, we can calculate how much distance the ball traveled in that time.
First, let´s calculate the distance traveled in the eastward direction:
x = x0 + v0x · t + 1/2 · ax · t² (x0 = 0, ax = 0 there is no eastward acceleration)
x = 50 ft/s · 3 s
x = 150 ft
And now let´s calculate the distance traveled in southward direction:
y = y0 + v0y · t + 1/2 · ay · t² (y0 = 0 and v0y = 0, initially, the ball does not have a southward velocity).
y = 1/2 · ay · t²
y = 1/2 · (-8 ft/s²) · (3 s)²
y = -36 ft
Then, the final position vector will be:
r = (150 ft, -36 ft, 0)
The traveled distance is the magnitude of the position vector:
To calculate the angle, we have to use trigonometry (see attached figure):
cos angle = adjacent side / hypotenuse
cos α = x/r
cos α = 150 ft / 154.3 ft
α = 13.6°
The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.
