More answers

Q: Ball kicked into the air at an angle of 34.0 degrees above the horizontal the initial velocity of the ball is 25.0 ms How long is the soccer ball in the air?

Write your answer...

Submit

Still have questions?

Continue Learning about Physics

To find the horizontal distance traveled by the soccer ball, you can use the equation: horizontal distance = horizontal velocity x time. The horizontal velocity is given by the formula Vx = V0 cosθ, where V0 is the initial velocity and θ is the angle of projection. Substituting the given values: Vx = 10.0 m/s * cos(30°) = 8.66 m/s. Then, the horizontal distance = 8.66 m/s * 3.2 s = 27.71 meters.

Factors that can affect the value of the horizontal velocity of a ball include the initial speed at which the ball was thrown or kicked, the angle at which it was launched, air resistance, and any external forces acting on the ball such as friction or gravity.

The initial velocity can be found using the kinematic equation: (d = v_0t + \frac{1}{2}at^2), where (d = 32m), (a = -9.81 m/s^2) (acceleration due to gravity), and (t) can be calculated using the time it takes for the rock to fall from a height of 450m. The initial velocity (v_0) is the horizontal component of velocity; therefore, it is the found by (v_0 = \frac{d}{t}).

To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.

Problem: A football is kicked from the ground with an initial velocity of 20 m/s at an angle of 45 degrees above the horizontal. Determine the maximum height reached by the football. Answer: The maximum height can be found using the equation: H_max = (v^2 * sin^2(theta)) / (2g), where v is the initial velocity (20 m/s), theta is the launch angle (45 degrees), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in these values, the maximum height is calculated to be approximately 10.1 meters.

Related questions

Initial velocity is 10 m/s in the direction it was kicked. Final velocity is 0, when friction and air resistance finally causes it to come to a halt.

To find the horizontal distance traveled by the soccer ball, you can use the equation: horizontal distance = horizontal velocity x time. The horizontal velocity is given by the formula Vx = V0 cosθ, where V0 is the initial velocity and θ is the angle of projection. Substituting the given values: Vx = 10.0 m/s * cos(30°) = 8.66 m/s. Then, the horizontal distance = 8.66 m/s * 3.2 s = 27.71 meters.

Factors that can affect the value of the horizontal velocity of a ball include the initial speed at which the ball was thrown or kicked, the angle at which it was launched, air resistance, and any external forces acting on the ball such as friction or gravity.

The initial velocity can be found using the kinematic equation: (d = v_0t + \frac{1}{2}at^2), where (d = 32m), (a = -9.81 m/s^2) (acceleration due to gravity), and (t) can be calculated using the time it takes for the rock to fall from a height of 450m. The initial velocity (v_0) is the horizontal component of velocity; therefore, it is the found by (v_0 = \frac{d}{t}).

To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.

Problem: A football is kicked from the ground with an initial velocity of 20 m/s at an angle of 45 degrees above the horizontal. Determine the maximum height reached by the football. Answer: The maximum height can be found using the equation: H_max = (v^2 * sin^2(theta)) / (2g), where v is the initial velocity (20 m/s), theta is the launch angle (45 degrees), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in these values, the maximum height is calculated to be approximately 10.1 meters.

When a ball is kicked at an angle, there is no acceleration along the horizontal direction (since there isn't any force along the direction ,ignoring viscous forces), so , its velocity along the horizontal direction remains unchanged.... according to the 1st law , velocity changes only when a net resultant force is applied on the ball , so , Newton's law is valid. only the initial angle of kick and the vertical component of velocity are mainly responsible for the distance travelled by the ball horizontally....

At the top of it path or anywhere else the Earth's gravity accelerates the ball downward. Regardless what the horizontal mothion is (disregarding air resistance), there are no accelerations after the kick. A ball kicked straight up to the same height will hit the ground at the same time as a ball kicked at 45 degrees.

The velocity of a football when it is hit can vary depending on factors such as the force of the impact and the angle at which it is hit. The initial velocity will be determined by the force exerted on the ball, and it will gradually decrease due to air resistance and other factors once it is in motion.

When it's at its maximum height its speed will be zero.

I assume the question is referring to a ball say that is being kicked, in this case some fairly simplistic physics determines that 45 degrees is the perfect angle, this calculation relies upon the fact that the ground is at the same height when it was kicked as when to when it comes to rest. (Delta Yx =0. Hence a level ground) However... If you are talking about a shot put for instance there is a different story, the optimum angle for this type of throw would be around 42 degrees. This value is different from the first due to the differential in heights between Y(x0) and Y(xfinal) where X is distance, Y states the vertical plane and 0 when t=0 (initial height) and then final when Vx&Vy=0 (aka rest). A key point to remember in the proof of this is that velocity, time and distance are all independent of mass. Regardless of how you choose to calculate the initial velocity the explanation stated above is always true for the angle both in a vacuum and in a real world environment. You didn't ask for a scientific proof so in summary: 45 degrees for a ground object (eg. football/soccer) 42 degrees for an object thrown from torso height (eg. shotput) - Approximation

The acceleration in the vertical direction is due to gravity and is approximately 9.8 m/s^2 downward. The vertical acceleration remains constant throughout the ball's flight trajectory.