A soccer ball is kicked from the ground. After travelling a horizontal distance of 25m, it just passes over a 2m tall fence before hitting the ground at a horizontal distance of 35m.
a) Determine an equation to represent the height of the ball relative to its horizontal distance away from its starting point(origin).
b) What is the maximum height of the ball?
To determine an equation representing the height of the ball relative to its horizontal distance, let's analyze the given information.
Let's assume that the initial height of the ball when it is kicked is 0m (ground level). We are given that the ball passes over a 2m tall fence and lands on the ground.
Let's denote the height of the ball above the ground as "h" and the horizontal distance travelled by the ball as "d".
a) To determine the equation, we need to find the relationship between the variables h and d. We can use the concepts of projectile motion and the kinematic equations.
At the highest point of its trajectory, the ball will be at its maximum height. At this point, the vertical velocity (Vy) is 0 because the ball momentarily stops moving upwards before falling back down.
Using the equation for vertical displacement (h) in terms of initial vertical velocity (Vy0), time (t), and acceleration due to gravity (g):
h = Vy0 * t - (1/2) * g * t^2
Since the initial vertical velocity is 0 at the highest point, the equation simplifies to:
h = - (1/2) * g * t^2
We can also find the time (t) it takes for the ball to reach the maximum height by using the equation for vertical velocity (Vy) in terms of initial vertical velocity (Vy0), acceleration due to gravity (g), and time (t):
Vy = Vy0 - g * t
Since at the maximum height, the vertical velocity is 0, the equation becomes:
0 = Vy0 - g * t
Solving for t, we get:
t = Vy0 / g
Now, let's find Vy0. We know the distance the ball travels horizontally before hitting the ground again is 35m.
Using the equation for horizontal displacement (d) in terms of initial horizontal velocity (Vx0) and time (t):
d = Vx0 * t
Since the horizontal velocity (Vx) remains constant throughout the motion, we can rewrite the equation as:
d = Vx * t
Now, let's find Vx0. We know that the ball travels a horizontal distance of 25m before passing over the fence. Therefore:
25 = Vx0 * t
Substituting t = Vy0 / g from the equation above, we get:
25 = Vx0 * (Vy0 / g)
Simplifying further, we have:
25 = (Vx0 * Vy0) / g
Rearranging the equation, we get:
Vx0 * Vy0 = 25 * g
Since the horizontal and vertical velocities are related, we can express Vx0 as a function of Vy0:
Vx0 = (25 * g) / Vy0
Now, substituting Vx0 = d / t and t = Vy0 / g in the above equation, we can rewrite it as:
d = (25 * g * Vy0) / (Vy0 * g)
Canceling out Vy0 and g in the equation, we get:
d = 25
Therefore, the equation representing the height of the ball relative to its horizontal distance is:
h = - (1/2) * g * (d/25)^2
b) To determine the maximum height of the ball, we substitute the horizontal distance (d) at the highest point of the trajectory into the equation:
h = - (1/2) * g * (35/25)^2
Calculating this, we find the maximum height of the ball.
You know that y(0) = y(35) = 0
So, y = ax(x-35)
Since y(25) = 2,
a(25)(25-35) = 2
a = -1/125
y(x) = -1/125 x(x-35)
max height is achieved at x = 35/2, so
y(35/2) = 49/20 = 2.45