A ball rolls off the edge of a balcony at 10 m/s. what is the speed after 1s in the air?
vertical: vf=gt=9.8m/s
horizontal: v=10
speed=sqrt(Vf^2+v^2)
To find the speed of the ball after 1 second in the air, we can use the formula for vertical motion under gravity:
Vf = Vi + gt
where:
- Vf is the final velocity (speed) after 1 second in the air,
- Vi is the initial velocity (speed) at which the ball rolls off the edge of the balcony (which is 10 m/s),
- g is the acceleration due to gravity (which is approximately 9.8 m/s^2),
- t is the time in seconds (which is 1 second in this case).
Now let's substitute the values into the formula and calculate the final velocity:
Vf = 10 m/s + (9.8 m/s^2)(1 s)
Vf = 10 m/s + 9.8 m/s^2
Vf = 19.8 m/s
Therefore, the speed of the ball after 1 second in the air is approximately 19.8 m/s.
To determine the speed of the ball after 1 second in the air, we need to consider the motion of the ball. When the ball rolls off the edge of the balcony, it enters free fall and accelerates due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2, assuming no air resistance.
To find the speed after 1 second, we can use the equation:
v = u + at
Where:
v is the final velocity (speed) of the ball
u is the initial velocity (speed) of the ball
a is the acceleration
t is the time
In this case:
u = 10 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity)
t = 1 s (time)
Plugging the values into the equation, we have:
v = 10 m/s + (9.8 m/s^2)(1 s)
v = 10 m/s + 9.8 m/s^2
v = 19.8 m/s
Therefore, the speed of the ball after 1 second in the air is 19.8 m/s.