Answer:
We can use the kinematic equations to solve this problem. Here are the steps:
(a) To find the initial velocity of the ball:
Let v be the initial velocity of the ball, and g be the acceleration due to gravity (-9.81 m/s^2).
The horizontal component of the velocity will remain constant throughout the motion, so we can write:
vx = v cos(45°)
where vx is the horizontal component of the velocity.
At the highest point, the vertical velocity of the ball will be zero, so we can use the equation:
vy^2 = u^2 + 2gh
where u is the initial vertical velocity, and h is the maximum height. We know that u = v sin(45°), so we can write:
(v sin(45°))^2 = v^2 + 2gh
Substituting the given values, we get:
(v^2)/2 = 2.4 m
v = √(4.8g) ≈ 9.25 m/s
Therefore, the initial speed of the ball was approximately 9.25 m/s.
(b) To find the time taken to reach the greatest height:
We can use the kinematic equation:
v = u + gt
At the highest point, the vertical velocity of the ball is zero, so we can write:
0 = (v sin(45°)) + gt
Solving for t, we get:
t = (v sin(45°)) / g
Substituting the value of v, we get:
t = (9.25 m/s) / (sqrt(2) * 9.81 m/s^2) ≈ 0.66 s
Therefore, the time taken to reach the greatest height was approximately 0.66 seconds.
(c) To find the horizontal distance between the point of kick and the foot of the goal post:
We can use the kinematic equation:
d = vxt
where d is the horizontal distance, and t is the time taken to reach the foot of the goal post. We already know the time t from part (b), so we just need to calculate the horizontal velocity:
vx = v cos(45°) ≈ 6.54 m/s
Substituting the values, we get:
d = (6.54 m/s) x (0.66 s) ≈ 4.32 m
Therefore, the horizontal distance between the point of kick and the foot of the goal post was approximately 4.32 meters.