A ball rolls off the edge of a balcony at 8.0 m/s. Find its speed after 3.8 s in the air. Use g = 10. m/s2
Vi=0, t=3.8s, a=g=10m/s^2, and Vf=?. Use Vf=Vi +at, which turns into Vf= (10m/s^2)(3.8s). Initial velocity in the y-direction is 0.
v(x0) =v(x)=8 m/s
v(y0) =0
v(y)=gt=10•3.8=38 m/s
v=sqrt{v(x)² +v(y)²} =
=sqrt (8²+38²) = 38.8 m/s
An autographed baseball rolls off of a 1.1 m
high desk and strikes the floor 0.15 m away
from the desk.
How fast was it rolling on the desk before
it fell off? The acceleration of gravity is
9.81 m/s
2
.
Answer in units of m/s
To find the speed of the ball after 3.8 seconds in the air, we can use the equations of motion. In this case, we have the initial velocity (u) of the ball, which is 8.0 m/s, the time (t) in the air, which is 3.8 seconds, and the acceleration due to gravity (g), which is 10 m/s^2.
The equation we can use is:
v = u + gt
Where:
- v is the final velocity of the ball
- u is the initial velocity of the ball
- g is the acceleration due to gravity
- t is the time
Substituting the given values into the equation, we have:
v = 8.0 m/s + (10. m/s^2)(3.8 s)
v = 8.0 m/s + 38 m/s
v = 46.0 m/s
Therefore, the speed of the ball after 3.8 seconds in the air is 46.0 m/s.