Question

A soccer ball is kicked with a speed of 9.25 m/s at an angle of 30.0 degrees above the horizontal.

If the ball lands at the same level from which it was kicked, how long was it in the air?

Answer 1

Concepts and reason

The concepts required to solve the problem are the fundamental kinematic equations for the projectile motion and the component representation of a vector.

Determine the horizontal and vertical components of initial velocity of the projectile using the vector components of velocity. Derive the equation for time taken to the projectile motion of the soccer ball. Substitute the velocity corresponds to the displacement and calculate the time of flight of the ball.

Fundamentals

Velocity of an object is the change in velocity. Velocity is vector quantity. The projectile launched at an angle has two perpendicular components of velocity. The two dimensional motion of the projectile is possible with the component of velocity. Only vertical velocity will displace the projectile vertically and the horizontal velocity keeps the horizontal motion of the projectile. The horizontal component of velocity is,

${u_x} = u\cos \theta$

Here, ${u_x}$ is the horizontal component of velocity, $u$ is the magnitude of velocity and $\theta$ is the angle of projection of the velocity vector from the horizontal axis. The horizontal velocity will be constant throughout the transit of the projectile.

The vertical component of the velocity vector is,

${u_y} = u\cos \theta$

Here, ${u_y}$ is the vertical component of velocity.

The time of flight of a projectile represents the time taken by the projectile to complete the motion from initial launching point to the final landing point. The time can be derived from the time dependent kinematic equation for displacement of projectiles. The kinematic equation for vertical displacement of projectile motion is,

$s = {u_y}t - \frac{1}{2}g{t^2}$

Here, $s$ is the displacement, $t$ is the time taken for the motion, $g$ is the acceleration due to gravity. The negative sigh is the sign of acceleration due to gravity. The acceleration due to gravity acts always downwards. Since the projectile moves against the acceleration, the sign of acceleration due to gravity is negative in projectile motion.

The projectile launches and lands at the same level have the vertical displacement zero at the end of the motion. The vertical displacement of the projectile is,

${u_y}t - \frac{1}{2}gt = 0$

The time of flight of the projectile is,

$t = \frac{{2{u_y}}}{g}$

The vertical component of initial velocity of the soccer ball is,

${u_y} = u\sin \theta$

Here, ${u_y}$ is the vertical component of initial velocity of the soccer ball, $u$ is the magnitude of initial velocity of the soccer ball and $\theta$ is the angle of projection of the soccer ball.

Substitute 9.25 m/s for $u$ and $30.0^\circ$ for $\theta$ to calculate the vertical component of initial velocity. The vertical component of initial velocity of the soccer ball is,

$\begin{array}{l}\\{u_y} = \left( {9.25\,{\rm{m/s}}} \right)\left( {\sin 30.0^\circ } \right)\\\\{u_y} = 4.62\,{\rm{m/s}}\\\end{array}$

The time of flight of the soccer ball is,

$t = \frac{{2{u_y}}}{g}$

Substitute the acceleration due to gravity, g as 9.81 m/s² and the vertical component of the velocity u_y as 4.625 m/s in the equation for time of flight.

$\begin{array}{l}\\t = \frac{{2\left( {4.62\,{\rm{m/s}}} \right)}}{{9.81\,{\rm{m/}}{{\rm{s}}^2}}}\\\\t = 0.94\,{\rm{s}}\\\end{array}$

Ans:

The ball will be at air for $0.94{\rm{ s}}$ .