Question

A soccer player kicked a ball with an initial velocity of 30 m/s at an angle of 45 degrees with horizontal. Calculate the time of flight, range, and the height of the ball. Show your work.

Answer 1

To solve the problem, we can break down the initial velocity into its horizontal and vertical components. Using the equations of projectile motion, we can find the time of flight, range, and height of the soccer ball. The time of flight (t) is √(2y/9.8), the range (x) is 30 * √(2y/9.8), and the height (y) is (1/2)at².

Step-by-step explanation:

To solve this problem, we can use the equations of projectile motion. First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, which is 30 m/s. The vertical component can be found using the equation v = v0 + at, where v is the final vertical velocity, v0 is the initial vertical velocity, a is the acceleration (which is due to gravity and is approximately -9.8 m/s²), and t is the time. Since the soccer ball is kicked from the ground, the initial vertical velocity is 0, so we have v = -9.8t.

Next, we can use the equation y = v0t + (1/2)at² to find the height (y) of the ball. Again, since the initial vertical velocity is 0, the equation reduces to y = (1/2)at². Plugging in the value of the acceleration (-9.8 m/s²) and solving for t, we get t = √(2y/9.8).

Finally, we can use the equation x = v0x * t to find the range (x) of the ball. Plugging in the values of v0x (which is the horizontal component of the initial velocity, which remains constant throughout the motion) and t (which we found earlier), we get x = 30 * √(2y/9.8).

To summarize, the time of flight (t) is √(2y/9.8), the range (x) is 30 * √(2y/9.8), and the height (y) is (1/2)at².

Answer 2

Step-by-step explanation:

It is given that,

Initial velocity of the soccer player, u = 30 m/s

It is launched at an angle of 45 degrees with the horizontal.

1. Time of flight :

The time taken by the projectile in the air and to comes to ground is called the time of flight. It is given by :

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

`T=(2(30)\ sin(45))/(9.8)`

T = 4.32 seconds

2. Range of the projectile

The horizontal distance covered by the projectile is called its range. It is given by :

`R=(u^2\ sin2\theta)/(g)`

`R=((30)^2\ sin2(45))/(9.8)`

R = 91.83 meters

3. Maximum height reached by the ball is given by :

`H=((u\ sin\theta)^2)/(2g)`

`H=((30* \ sin(45))^2)/(2* 9.8)`

H = 22.95 meters

Hence, this is the required solution.