A soccer ball is kicked with an initial horizontal velocity of 12 m/s and an initial vertical velocity of 18 m/s.

Question

A soccer ball is kicked with an initial horizontal velocity of 12 m/s and an initial vertical velocity of 18 m/s.

1)What is the initial speed of the ball?

2)What is the initial angle θ of the ball with respect to the ground?

3)What is the maximum height the ball goes above the ground?

4)How far from where it was kicked will the ball land?

5)What is the speed of the ball 2.9 seconds after it was kicked?

6)How high above the ground is the ball 2.9 seconds after it is kicked?

Answer 1

Hi Davinder

1. The initial speed of the ball when it is kicked is the magnitude of the vector created from the horizontal and vertical velocities. Thus S = sqrt(12^2 + 18^2)
2. The tangent of the initial angle is equal to the vertical velocity / horizontal velocity . Thus Theta is equal to the arctan of 18/12
3. The maximum height occurs where the vertical velocity reaches zero. You can use this equation V(final) ^2 = V(initial)^2 - 2 g H, where gis the acceleration of gravity and H is the height attained
4. The range R is equal to S^2 * sin (2*theta) You calculated S in question 1 and theta in question 2
5. The horizontal velocity is simply the initial velocity - it doesn't change in this problem The vertical velocity is the initial velocity (18) minus g (gravity) times t Thus 18-g(2.9). The net velocity is the pythagorean resolution of v(H) and v(v) - same calculation as in #1
6. The height depends only on initial vertical velocity , The equation is d= V(init) *t + 1/2 * g* t^2. plug in 2.9 seconds and crank it out.

Jim