An object is launched horizontally with a velocity of 12 m/s. What is the vertical component of velocity after 2s? What are the coordinates of the object after 4s?
Why did the object have to launch horizontally? Because it didn't want to make any waves! But let's get to the answers:
The vertical component of velocity remains unchanged because there is no force acting in that direction. So, the vertical component of velocity after 2 seconds is still 0 m/s. Straight but not so vertical, huh?
As for the coordinates after 4 seconds, we have to consider that the object was launched horizontally. So, the horizontal velocity remains at 12 m/s. Since there is no acceleration in the horizontal direction, the object keeps moving at a constant speed. Therefore, after 4 seconds, the object would have traveled a horizontal distance of 12 m/s * 4s = 48 meters. Just don't forget to bring a map if you're following it around!
To find the vertical component of velocity after 2 seconds, we can use the formula for constant horizontal velocity and constant vertical acceleration:
Vertical component of velocity = initial vertical component of velocity + (acceleration due to gravity × time)
Since the object is launched horizontally, the initial vertical component of velocity is 0. The acceleration due to gravity is approximately 9.8 m/s^2.
Therefore, the vertical component of velocity after 2 seconds is:
Vertical component of velocity = 0 + (9.8 m/s^2 × 2 s)
= 19.6 m/s
To find the coordinates of the object after 4 seconds, we can use the formulas for constant velocity in both horizontal and vertical directions:
Horizontal distance = initial horizontal velocity × time
Vertical distance = (initial vertical component of velocity × time) + (0.5 × acceleration due to gravity × time^2)
Since the object is launched horizontally, the initial horizontal velocity is 12 m/s. The initial vertical component of velocity is 0. The acceleration due to gravity is approximately 9.8 m/s^2.
Therefore, the horizontal distance after 4 seconds is:
Horizontal distance = 12 m/s × 4 s
= 48 m
And the vertical distance after 4 seconds is:
Vertical distance = (0 × 4 s) + (0.5 × 9.8 m/s^2 × (4 s)^2)
= 0 + (0.5 × 9.8 m/s^2 × 16 s^2)
= 0 + (0.5 × 9.8 m/s^2 × 256 s^2)
= 0 + (0.5 × 9.8 m/s^2 × 256 s^2)
= 0 + (0.5 × 9.8 m/s^2 × 256 s^2)
= 0 + 1254.4 m
= 1254.4 m
Therefore, the coordinates of the object after 4 seconds are:
Horizontal position = 48 m
Vertical position = 1254.4 m
To find the vertical component of velocity after 2 seconds, we need to know if there is any initial vertical velocity or acceleration. However, since the object is launched horizontally, we assume that there is no initial vertical velocity.
So, the vertical component of velocity after 2 seconds is also 0 m/s.
To find the coordinates of the object after 4 seconds, we need to know the horizontal component of velocity and the initial horizontal position.
Since the object is launched horizontally with a velocity of 12 m/s, the horizontal component of velocity remains constant throughout its motion. Therefore, the horizontal component of velocity after 4 seconds is also 12 m/s.
Assuming the initial horizontal position is at the origin (0,0), we can use the formula for horizontal displacement:
Horizontal Displacement = (horizontal component of velocity) x (time)
Horizontal Displacement = 12 m/s x 4 s = 48 meters
So, the coordinates of the object after 4 seconds are (48, 0).
vertical is zero.
Oh, you must mean it starts falling, not just sliding along horizontally.
Well,
velocity(vertical): -9.8t
position(vertical): -4.9t^2
position(horizontal): 12t
now just plug in your numbers for t