A local football player is practicing football on top of 25-m high hill. If he kicks the soccer ball with an initial horizontal velocity of 25 m/s,
a. what is its Vx or horizontal velocity?
b. after how many seconds will it land?
c. how many meters will be its range or horizontal displacement?
To answer these questions, we need to analyze the motion of the soccer ball. Assuming no air resistance, the only force acting on the ball in the horizontal direction is its initial velocity. In the vertical direction, the ball experiences the force of gravity, which causes it to accelerate downward.
a. To determine the horizontal velocity (Vx), we use the equation:
Vx = initial horizontal velocity = 25 m/s
b. To find the time it takes for the ball to land, we need to consider the vertical motion. The vertical distance the ball must fall is equal to the height of the hill, which is 25 m. The equation for the vertical motion under gravity is given by:
h = vi * t + (1/2) * g * t^2
where h is the vertical distance, vi is the initial vertical velocity (0 m/s as the ball is just being kicked horizontally), t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values:
25 = 0 * t + (1/2) * 9.8 * t^2
Simplifying:
4.9 * t^2 = 25
t^2 = 25 / 4.9
Taking the square root of both sides, we get:
t ≈ 2.04 s
So, the ball will land after approximately 2.04 seconds.
c. To find the horizontal displacement or range, we use the equation:
Range = horizontal velocity * time
Range = 25 m/s * 2.04 s
Range ≈ 51 meters
Therefore, the horizontal displacement or range of the ball will be approximately 51 meters.