A soccer ball is kicked with a speed of 16.7 m/s at an angle of 45.0 ∘ above the horizontal.If the ball lands at the same level from which it was kicked, for what amount of time was it in the air?
2 * vertical velocity / g
16.7 * √2 / 9.81
To find the amount of time the ball was in the air, we can use the vertical motion equations.
Step 1: Resolve the initial velocity into its horizontal and vertical components:
The initial velocity of the ball can be resolved into two components: the horizontal component (Vx) and the vertical component (Vy).
Given:
Initial speed = 16.7 m/s
Launch angle = 45.0 degrees above the horizontal
The horizontal component of the velocity (Vx) remains constant throughout the motion and is given by:
Vx = Initial speed * cos(angle)
Substituting the given values:
Vx = 16.7 m/s * cos(45.0 degrees)
Step 2: Determine the time taken for the vertical motion:
The vertical component of the velocity (Vy) changes due to gravity. The time taken for the ball to reach its highest point and return to the same vertical level is the same.
Using the vertical motion equation:
Vy = Initial speed * sin(angle) - (g * t)
Since the ball returns to the same vertical level, the final velocity in the vertical direction (Vy) is zero when the ball lands. Thus, we can set Vy to zero and solve for t.
0 = Initial speed * sin(angle) - (g * t)
Simplifying the equation:
t = Initial speed * sin(angle) / g
Substituting the given values:
t = 16.7 m/s * sin(45.0 degrees) / (9.8 m/s^2)
Calculating the value of t:
t = 1.20 seconds
Therefore, the ball was in the air for 1.20 seconds.