Projectile Motion Analysis: Where Will the Ball Land After a 21m/s Throw from 10m Height?

When analyzing the motion of an object thrown into the air, such as a ball, we need to consider both its horizontal and vertical components of motion. This article will walk through the analysis of a specific scenario where a ball is thrown from a height of 10 meters with an initial velocity of 21 meters per second. We will break down the motion into its components and use the relevant kinematic equations to determine where the ball lands on the ground.

Vertical Motion

The vertical motion of the ball is primarily influenced by gravity, which causes it to accelerate downwards. We can describe the vertical position over time using the following kinematic equation:

h h_0 v_{y0}t - frac{1}{2}gt^2

Where:

h is the final height h_0 is the initial height (10 meters) v_{y0} is the initial vertical velocity (0 m/s, assuming the throw is horizontal) g is the acceleration due to gravity (9.81 m/s2) t is the time of flight

When the ball hits the ground, h 0. Substituting the known values, we get:

0 10 0 - frac{1}{2} cdot 9.81 cdot t^2

Which simplifies to:

t^2 frac{10}{4.905} approx 2.04

Therefore, the time of flight t approx sqrt{2.04} approx 1.43 text{ seconds}.

Horizontal Motion

The horizontal motion of the ball is not affected by gravity, so it moves with a constant velocity. The horizontal distance L can be calculated as follows:

L v_x cdot t

Where:

L is the horizontal distance v_x is the initial horizontal velocity (21 m/s) t is the time of flight (1.43 seconds)

Therefore, the horizontal distance L 21 cdot 1.43 approx 30.03 text{ meters}.

Conclusion

The ball will hit the ground approximately 30.03 text{ meters} horizontally from the point where it was thrown. This calculation demonstrates the application of kinematic equations to analyze projectile motion, which is a fundamental concept in physics.

Additional Considerations

Given the problem statement, it is also important to clarify whether the ball was thrown downward or sidewise. In this case, the initial vertical velocity is 0 m/s, indicating that the throw is horizontal. Therefore, the ball was thrown horizontally, parallel to the ground, and it follows a parabolic trajectory.

Related Questions

The problem provides more detailed variables and equations to further analyze the scenario. If we denote:

L as the distance in meters where the ball will land T as the time in seconds taken by the ball to land V as the velocity in meters per second of the ball when it lands V_x as the horizontal component of velocity V_y as the vertical component of velocity

Based on the assumptions:

The ball was thrown horizontally parallel to the ground. The ball follows a 2D motion following a parabolic trajectory.

Using the following kinematic relations:

0 - frac{1}{2}gt^2 10 quad [Equation 1a]

L v_xt quad [Equation 1b]

V_x v_{x0} quad [Equation 1c]

V_y -gt quad [Equation 1d]

V sqrt{V_x^2 V_y^2} quad [Equation 1e]

From Equation 1a, we find:

frac{1}{2}9.8t^2 10 quad [g 9.8 m/s^2]

t^2 frac{10}{4.9} approx 2.04

t sqrt{2.04} approx 1.43 text{ seconds}

From Equation 1d and the calculated t, we get:

V_y -9.8 cdot 1.43 -14 text{ m/s}

From Equation 1e, 1c, and 1d, we find:

V sqrt{21^2 (-14)^2} sqrt{441 196} sqrt{637} 25.24 text{ m/s}

From Equation 1b and the calculated t, we find:

L 21 cdot 1.43 30 text{ meters}

Therefore, the ball will hit the ground approximately 30 text{ meters} horizontally from the point where it was thrown.