The local football player is practicing football on top of a 25 m high hill. He kicks the soccer ball, which lands 40 m from the base of the hill. Determine the initial horizontal velocity of the ball. Round off to 1 decimal place.
Assuming a cliff, instead of a hill (the base of a hill could be some distance away), how long does it take to fall 25m?
4.9t^2 = 25
since the horizontal speed is constant,
v = 40/t
1.6
The initial horizontal velocity of the ball is approximately 1.6 m/s.
Why did the soccer ball go to college? Because it wanted to get a kick out of higher education!
To determine the initial horizontal velocity of the ball, we need to consider the horizontal motion of the ball. Since we know the height of the hill (25 m) and the horizontal displacement (40 m), we can use the formula:
Horizontal velocity (Vx) = horizontal displacement / time
Since there is no acceleration in the horizontal direction, the time taken to reach the ground will be the same as the time taken to travel horizontally.
To find the time, we can use the formula:
Vertical displacement = (1/2) * acceleration * time^2
Here, the initial vertical velocity is 0 m/s, the vertical displacement is -25 m (negative since the ball is going down), and the acceleration due to gravity is 9.8 m/s^2.
Substituting the values into the formula, we get:
-25 = (1/2) * 9.8 * time^2
Simplifying the equation, we have:
-25 = 4.9 * time^2
Dividing both sides by 4.9, we get:
-25 / 4.9 = time^2
Using a calculator, we find:
time ≈ 1.42857 seconds
Now, we can use this time to find the horizontal velocity:
Vx = 40 m / 1.42857 s ≈ 27.9 m/s
Therefore, the initial horizontal velocity of the ball is approximately 27.9 m/s.
To determine the initial horizontal velocity of the ball, we need to use the concept of projectile motion. In this case, the soccer ball moves in a curved path in two dimensions: horizontal and vertical.
The key thing to note is that the initial vertical velocity (Viy) of the ball is zero because it is kicked horizontally. This means that gravity only affects the ball's vertical motion while not impacting its horizontal motion.
Let's break down the problem using the following equations of motion:
1. Vertical displacement equation:
h = Vit + (1/2)gt^2
2. Horizontal displacement equation:
d = Vixt
where:
h = vertical displacement (height of the hill) = 25 m
Viy = initial vertical velocity = 0
g = acceleration due to gravity = 9.8 m/s^2
t = time of flight
d = horizontal displacement = 40 m
Vix = initial horizontal velocity (what we need to find)
Since Viy = 0, we can use the vertical displacement equation to solve for the time of flight (t):
25 m = (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation:
25 = 4.9t^2
Dividing by 4.9:
5.1 = t^2
Taking the square root:
t ≈ 2.26 seconds
Now that we have the time of flight (t), we can use the horizontal displacement equation to solve for the initial horizontal velocity (Vix):
40 m = Vix * 2.26 s
Rearranging the equation:
Vix = 40 m / 2.26 s
Calculating:
Vix ≈ 17.7 m/s
So, the initial horizontal velocity of the ball is approximately 17.7 m/s.