A soccer ball is kicked with a speed of 9.15 m/s at an angle of 55.0^\circ above the horizontal.
1-If the ball lands at the same level from which it was kicked, how long was it in the air?
You should be able to do this with my answer to the other question.
To find how long the soccer ball was in the air, we can use the equations of motion. The equation we will use is the one that relates the horizontal distance (range) traveled by an object to its initial velocity, launch angle, and time of flight.
The equation for the horizontal distance traveled, or range (R), is given by:
R = (v^2 * sin(2θ)) / g
where:
v = initial velocity of the ball (9.15 m/s)
θ = launch angle of the ball (55.0 degrees)
g = acceleration due to gravity (9.8 m/s^2)
We can rearrange the equation to solve for the time of flight (t):
t = (2 * v * sin(θ)) / g
Now we can substitute the given values into the equation to calculate the time of flight:
t = (2 * 9.15 * sin(55.0)) / 9.8
Calculating this expression will give us the time of flight of the soccer ball.