a soccer ball is kicked from the ground with an initial speed of 19.5 meter per second at upward of 45 degrees. a player 55 meter away in the direction of the kick starts running to meet the ball at the instant. what must be his average speed if he is catch the ball just before it hits the ground
It hits the ground
(Vo^2/g) = 38.8 meters away
The other player was
55-38.8 = 16.2 meters
away when it was kicked. He must run that distance before the ball hits the ground. That running time is
2*Vo*sin45/g = 2.814 seconds
Required average speed = 16.2/2.814 = 5.76 m/s
vertical velocity=19.5Cos45=13.79m/s. Time up=(v-u)/g=(0-13.79/-9.8=1.42s. T(up)+T(down)=2*1.42=2.85s. avrgS=D/T=55/2.85=19.30m/s.
To determine the average speed the player must have to catch the ball just before it hits the ground, we need to consider the vertical and horizontal components separately.
1. Vertical Motion:
The initial vertical speed can be calculated using the given initial speed and launch angle. We can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Assuming no air resistance and neglecting the effect of gravity over the relatively short time period, the final vertical velocity for the ball just before it hits the ground is zero.
viy = visinθ
viy = 19.5 * sin(45°)
viy ≈ 19.5 * 0.707 ≈ 13.77 m/s
2. Horizontal Motion:
Since there are no horizontal forces acting on the ball once it is kicked, the horizontal velocity remains constant throughout the motion. Therefore, the horizontal distance covered by the ball is constant.
Thus, the player needs to run a distance of 55 meters to meet the ball before it hits the ground.
Now we know the horizontal distance and the total time it takes for the ball to reach the ground. We can use the equation d = vt, where d is distance, v is velocity, and t is time to find the total time of flight.
55 = vix * t
For the horizontal motion, we can use the equation vix = vi * cosθ.
vix = 19.5 * cos(45°)
vix ≈ 19.5 * 0.707 ≈ 13.77 m/s
55 = 13.77 * t
t ≈ 55 / 13.77 ≈ 3.99 seconds
Now, we can find the average speed using the total distance covered and the total time taken.
Average speed = Total distance / Total time
Average speed = (55 + 55) / 3.99
Average speed ≈ 27.57 m/s
Therefore, the player must have an average speed of approximately 27.57 m/s to catch the ball just before it hits the ground.
To solve this problem, we need to break it down into two parts: horizontal motion and vertical motion.
First, let's analyze the horizontal motion. The player starts running 55 meters away from the initial position of the ball. Since there are no horizontal forces acting on the ball, it will continue moving horizontally with a constant speed throughout its flight. Therefore, the time it takes for the ball to reach the player is the same as the time it takes for the ball to fall vertically to the ground.
Next, let's analyze the vertical motion. The initial speed of the ball is 19.5 m/s, and it is kicked at an upward angle of 45 degrees. We can decompose this initial velocity into its vertical and horizontal components. The vertical component is given by: V_y = V_initial * sin(theta). Thus, V_y = 19.5 m/s * sin(45) = 13.8 m/s.
Using the vertical motion equations, we can find the time it takes for the ball to reach the ground. The equation for vertical displacement is given by: d_y = V_y * t - (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Since the ball starts at ground level (d_y = 0) and eventually reaches the ground, we can set the equation equal to zero:
0 = 13.8 m/s * t - (1/2) * (9.8 m/s^2) * t^2.
Simplifying this equation, we get -4.9 t^2 + 13.8 t = 0.
Factoring out t, we have t * (-4.9t + 13.8) = 0.
From this equation, we have two possible solutions: t = 0 (which is not physically possible in this case) and t = 2.82 seconds. Therefore, the time it takes for the ball to reach the ground is approximately 2.82 seconds.
Since the horizontal distance is 55 meters and the time is 2.82 seconds, we can calculate the average speed required for the player to catch the ball just before it hits the ground by dividing the distance by the time:
Average speed = distance / time = 55 meters / 2.82 seconds ≈ 19.50 m/s.
Therefore, the player must maintain an average speed of approximately 19.50 m/s to catch the ball just before it hits the ground.