if a soccer player kicks a ball at an angle of 36 degrees and it lands 31.4 meters away. what is the initial velocity of the ball

hf=hi+vivertical*timeinair- 9.8 time^2

0=0+V*sin36*t-9.8 t^2

so what is time in air.
timeinair= 31.4/Vhorizontal= 31.4/Vcos36

Vsin36*t=9.8t^2
Vsin36=9.8 t

V= 9.8/sin36 * 31.4/Vcos36

V= sqrt (9.8*31.4/sin36cos36)

To find the initial velocity of the ball, we can use the horizontal and vertical components of the velocity.

Given:
Angle of projection (θ) = 36 degrees
Horizontal distance (d) = 31.4 meters

The horizontal component of velocity (Vx) remains constant and is given by:

Vx = V * cos(θ)

where V is the initial velocity.

Since the ball lands after traveling a certain horizontal distance, we can set up an equation:

d = Vx * t

where t is the time of flight.

Now, let's calculate Vx:

Vx = V * cos(36)

Next, we'll calculate the vertical component of velocity (Vy):

Vy = V * sin(θ)

The time of flight (t) can be determined using the equation:

t = 2 * Vy / g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now, substitute the values and equations:

d = Vx * t
31.4 = V * cos(36) * (2 * Vy / g)

From this equation, we can solve for V (initial velocity).