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Prj11 Kicked soccer ball

A soccer ball is kicked at 30^o with an initial velocity of 12 m/s.

What is the range, maximum height and time of flight?

Diagram the problem and take the origin of an x-y coordinate system at the point where the ball is kicked.

The acceleration due to gravity is in the (negative) y-direction and there is no acceleration in the x-direction. The velocity components are as calculated in the insert above.

Now write the equations for acceleration, velocity and position in the two directions.

As with most kinematics problems calculate the time of flight first. The time of flight comes out of the equation for y, and is obtained by setting y equal to zero (ground) and solving for t.

The times are 0, the ball is just leaving the ground, and 6/4.9 = 1.2, the ball is back on the ground. The time of flight is 1.2 s.

The range is the value of x at 1.2 s or

The maximum height occurs when v_y is zero.

This produces a value for t just 1/2 of the time of flight. This is one of the symmetry properties of parabolas. The time for maximum height is one-half the time of flight for any parabolic motion.

The height at this point is from y = . . .