Question

a soccer ball is kicked with a speed of 15.6m/s at an angle of 32.5 degrees 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?

Answer 1

To find the amount of time the soccer ball was in the air, we can break the initial velocity of the ball into its horizontal and vertical components. The time in the air can be calculated using the vertical velocity component and the vertical distance traveled.

Step-by-step explanation:

To find the amount of time the soccer ball was in the air, we can break the initial velocity of the ball into its horizontal and vertical components. The horizontal component will remain constant throughout the motion, while the vertical component will change due to the force of gravity.

Using the initial velocity of 15.6 m/s and the angle of 32.5 degrees above the horizontal, we can find the vertical velocity component using trigonometry. The vertical velocity component is given by Vy = V * sin(θ), where Vy is the vertical velocity, V is the initial velocity, and θ is the angle above the horizontal.

With the vertical velocity component, we can calculate the time the ball was in the air by dividing the vertical distance it travels by the vertical velocity. Since the ball lands at the same level, the vertical distance can be assumed to be zero. Therefore, the time in the air is given by Δt = 2 * (Vy / g), where Δt is the time in the air, Vy is the vertical velocity component, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Answer 2

Time of flight, t = 1.71 seconds

Step-by-step explanation:

It is given that,

Speed of the soccer ball, u = 15.6 m/s

Angle made with the horizontal,
`\theta=32.5^(\circ)`

Let t is the time for which the ball was in the air. The time duration for which the ball is in air is called the time of flight of the projectile. It is given by :

`t=(2u\ sin\theta)/(g)`

`t=(2(15.6)\ sin(32.5))/(9.8)`

t = 1.71 seconds

So, the ball was in the air for 1.71 seconds. Hence, this is the required solution.