A football player hits a ball at a speed of 8 meters per second with an angle of 30° from the ground. The ball strikes the ground at the point which is labeled with a point of projection. After impact with the ground, the ball bounces and the horizontal component of velocity remains the same but the vertical component is reversed in direction and curved in magnitude. The player running after the ball kicks it again at the point which is the horizontal distance of 1 m from the point where it bounced so the ball continues to the same direction. 1. Find the horizontal distance to show the point of projection and the points at which the ball strikes the ground. 2. Find the time interval between the ball striking the ground and a player kicking it again. 3. Find the height of the bull above the ground when it is kicked again.(take g=10 meters per second squared.

Here's the solution to the problem: Given: Initial speed of the ball, v_0 = 8 \, m/s Angle of projection, = 30^ Acceleration due to gravity, g = 10 \, m/s^2 --- 1. Find the horizontal distance to show the point of projection and the points at which the ball strikes the ground. This refers to the horizontal range of the first flight segment, from the initial kick to the first bounce. Step 1: Calculate the horizontal and vertical components of the initial velocity. v_x = v_0 = 8 (3