Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
and the time taken to cover the horizontal distance d is
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
The height of the crossbar is h = 3.05 m, so the ball passes
above the crossbar.
The resultant speed of the plane is (3) 226 m/s
Why?
We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).
So, calculating we have:
Hence, we have that the resultant speed of the plane is (3) 226 m/s
Have a nice day!
Explanation:What is centripetal acceleration?
Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.
Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.
The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration
a
c
a
c
a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.
1.15 m/s to the left (3 sig. fig.).
<h3>Explanation</h3>
Momentum is conserved between the two balls if they are not in contact with any other object. In other words,
, where
stands for mass and
stands for velocity, which can take negative values.
Let the velocity of objects moving to the right be positive.
,
.
Before the two balls collide:
,
.
After the two balls collide:
needs to be found, 
.
Again,
,
.
.
is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.
Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.
