Question

A football player punts a ball at an angle of 45° at 1.5 feet off the ground, with an initial velocity of 78 feet per second. how far away does the ball land, in feet, when hitting the ground? 207.295 feet 191.613 feet 170.892 feet 153.226 feet

Answer 1

The ball lands at a distance of 0 feet when hitting the ground.

Step-by-step explanation:

To find the distance that the ball lands, we need to analyze the horizontal component of the ball's motion. Since the ball is being kicked at an angle of 45°, the horizontal and vertical components of its initial velocity can be calculated using trigonometry.

The horizontal component of the initial velocity can be found using the formula: Vx = V * cos(theta)

where Vx is the horizontal component of the initial velocity, V is the magnitude of the initial velocity, and theta is the angle of projection.

In this case, V = 78 feet per second and theta = 45°, so:

Vx = 78 * cos(45°) = 78 * 0.7071 = 55.0566 feet per second

Now, we can use the horizontal component of the initial velocity to find the time of flight using the formula: t = 2 * Vy / g

where t is the time of flight, Vy is the vertical component of the initial velocity, and g is the acceleration due to gravity (assumed to be 32.2 feet per second squared).

Since the ball is being kicked with an initial vertical velocity of 0, Vy = 0. Therefore, the time of flight is:

t = 2 * 0 / 32.2 = 0 seconds

Since the ball is being kicked horizontally, the time of flight is 0 and the horizontal distance traveled can be found using the formula: d = Vx * t

where d is the horizontal distance traveled and Vx is the horizontal component of the initial velocity.

In this case, t = 0 and Vx = 55.0566 feet per second, so:

d = 55.0566 * 0 = 0 feet

Therefore, the ball lands at a distance of 0 feet when hitting the ground.

Answer 2

The ball will land approximately 191.613 feet away from the punt.

Step-by-step explanation:

To find the distance the ball lands, we can separate the motion into horizontal and vertical components. Since the angle of the initial velocity is 45°, the horizontal component of the velocity is equal to the vertical component of the velocity.

The initial vertical velocity can be found using the formula: Vy = V * sin(angle), where V is the initial velocity and angle is the launch angle. In this case, Vy = 78 * sin(45°) = 55.355 ft/s.

The time it takes for the ball to reach the ground can be found using the formula: t = 2 * Vy / g, where g is the acceleration due to gravity (32.2 ft/s^2). Plugging in the values, we get t = 2 * 55.355 / 32.2 = 3.44 s.

The horizontal distance traveled by the ball can be calculated using the formula: Dx = Vx * t, where Vx is the horizontal component of the velocity and t is the time. Since Vx = Vy, the horizontal distance traveled is Dx = 55.355 * 3.44 = 191.613 feet.