a ball is kicked with a velocity of 12 m/s at an angle of 42 degress above the horizontal. how far away does the ball land?

R = range = (V^2/g)*sin(2A)

2A = 84 degrees; V = 12 m/s, g = 9.81 m/s^2

Solve for R

See if you can derive the formula. It used the fact that 2sinA*cos A = sin(2A)

1,A foot ball is kicked at avelocity of 15m/s at angel of 25 degeres to the horizontal. calculate


a, the total flight time

b, the maximum height

c the range of the ball

To find the distance the ball lands, we need to break down the initial velocity of the ball into its horizontal and vertical components.

The horizontal component of the velocity can be found using the formula Vx = V * cos(θ), where V is the initial velocity and θ is the angle of projection.

Vx = 12 m/s * cos(42°)
Vx ≈ 12 m/s * 0.7431
Vx ≈ 8.917 m/s

The vertical component of the velocity can be found using the formula Vy = V * sin(θ), where V is the initial velocity and θ is the angle of projection.

Vy = 12 m/s * sin(42°)
Vy ≈ 12 m/s * 0.6691
Vy ≈ 8.029 m/s

Since the horizontal motion is unaffected by gravity, the time of flight can be determined using the vertical component of velocity:

t = (2 * Vy) / g
t = (2 * 8.029 m/s) / 9.8 m/s²
t ≈ 1.64 s

Where g is the acceleration due to gravity (approximately 9.8 m/s²).

The distance the ball lands can be calculated using the horizontal component of velocity and the time of flight:

distance = Vx * t
distance = 8.917 m/s * 1.64 s
distance ≈ 14.6 meters

Therefore, the ball will land approximately 14.6 meters away.