A soccer player kicks the ball toward a goal that is 16,8m in front of him . The ball leaves his foot at a speed of 16,0m/s and an angle 28,0degree above the ground.Find the speed of the ball when the goalie catches it in front of the net?
vx stays constant at 16cos28
time to goal = 16.8/16cos28
vy at goal = = 16.8sin28(t)- 1/2gt^2
Then Pythagoean with vx and vy
To find the speed of the ball when the goalie catches it, we can use the principles of projectile motion.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) remains constant at all times, while the vertical component (Vy) changes due to the effect of gravity.
Given:
Initial velocity (V₀) = 16.0 m/s
Launch angle (θ) = 28.0 degrees
Distance to the goal (x) = 16.8 m
Acceleration due to gravity (g) = 9.8 m/s²
Step 1: Calculate the horizontal component of the initial velocity (Vx).
Vx = V₀ * cos(θ)
Vx = 16.0 m/s * cos(28.0°)
Vx ≈ 14.37 m/s
Step 2: Calculate the time it takes for the ball to reach the goal.
We can use the distance formula for horizontal motion: x = Vx * t.
16.8 m = 14.37 m/s * t
t ≈ 1.17 s
Step 3: Calculate the vertical component of the velocity at that time (Vy).
Using the time calculated in Step 2, we can use the kinematic equation for vertical motion: y = V₀y * t + 0.5 * g * t².
Since the ball leaves the foot at ground level, the initial vertical velocity (V₀y) is zero.
0 = 0 + 0.5 * (-9.8 m/s²) * (1.17 s)²
Vy ≈ -6.14 m/s
Step 4: Calculate the final velocity of the ball when the goalie catches it.
The final velocity can be calculated using the Pythagorean theorem:
Vf = sqrt(Vx² + Vy²)
Vf = sqrt((14.37 m/s)² + (-6.14 m/s)²)
Vf ≈ 15.9 m/s
Therefore, the speed of the ball when the goalie catches it in front of the net is approximately 15.9 m/s.
ah --- if the altitude of the kicking foot and the goalie hands is the same, then the kinetic energy is the same.
same (1/2) m v^2
so same 16 m/s
(ignoring air friction)