Answer:
72.53 mi/hr
Explanation:
From the question given above, the following data were obtained:
Vertical distance i.e Height (h) = 8.26 m
Horizontal distance (s) = 42.1 m
Horizontal velocity (u) =?
Next, we shall determine the time taken for the car to get to the ground.
This can be obtained as follow:
Height (h) = 8.26 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
8.26 = ½ × 9.8 × t²
8.26 = 4.9 × t²
Divide both side by 4.9
t² = 8.26 / 4.9
Take the square root of both side by
t = √(8.26 / 4.9)
t = 1.3 s
Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:
Horizontal distance (s) = 42.1 m
Time (t) = 1.3 s
Horizontal velocity (u) =?
s = ut
42.1 = u × 1.3
Divide both side by 1.3
u = 42.1 / 1.3
u = 32.38 m/s
Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:
1 m/s = 2.24 mi/hr
Therefore,
32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s
32.38 m/s = 72.53 mi/hr
Thus, the car was moving at a speed of
72.53 mi/hr.
The answer is b. because they move freely
Answer:
Can you give me a better explantion
Explanation:
Magnitude = √(x-component)² + (y-component)²
Magnitude = √(-22.5m)² + (54.1m)²
Magnitude = √(506.25m² + 2926.81m²)
Magnitude = √(3,433.06 m²)
Magnitude = 58.6 meters
Direction = arctan(y-component/x-=component)
Direction = arctan(-54.1m/22.5m)
Direction = arctan(-2.404)
The angle between the direction of upper a overscript right-arrow endscripts and the positive direction of x is 67.4°.
That is, the vector upper a overscript right-arrow endscripts is in the 4th quadrant, a little below southeast, pointing from the origin in the direction of 67.4 degrees below the positive x-axis.