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.
States that particles are attracts with every other particle. wich force is directily proportional product of two masses and inversely proportional to the distance between the centers.
Diagram 4 is the correct answer.
Answer:
0.976 c
Explanation:
= velocity of object 1 relative to earth = 0.80 c
= velocity of object 2 relative to object 1 = 0.80 c
= velocity of object 2 relative to earth
Velocity of object 2 relative to earth is given as


= 0.976 c
Data !
hope this helped <3