A soccer play kicks the ball toward a goal that is 29m in front of him. The ball leaves his foot at a speed of 19m/s and at an angle of 32degrees above the ground. Find the speed of the ball when the goalie catches it in front of the net.
my work:
vxi=16.1
vxf=16.1
ax=0
LOL - trick question
energy is conserved
if the ball is caught at the same height as it left the foot, the speed is the same. 19 m/s
same horizontal component (of course)
same vertical component but down not up (conservation of energy)
To find the speed of the ball when the goalie catches it, we can use the equations of motion. We need to break down the initial velocity into its horizontal and vertical components.
First, let's find the initial velocity of the ball in the horizontal direction (vxi) using the given information. The angle of 32 degrees above the ground does not affect the horizontal velocity component.
vxi = 19 m/s * cos(32°) = 16.1 m/s
We can also find the initial velocity of the ball in the vertical direction (viy) using the given information.
viy = 19 m/s * sin(32°) ≈ 10.14 m/s
Now, we need to consider the horizontal motion of the ball. Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant throughout the motion.
vxf = vxi = 16.1 m/s
Next, let's look at the vertical motion of the ball. We can use the equation of motion in the vertical direction to find the time it takes for the ball to reach the goalie's position.
y = viyt - 0.5 * g * t^2
Where:
y is the vertical distance traveled by the ball (29 m)
viy is the initial vertical velocity (10.14 m/s)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time taken (unknown)
Rearranging the equation and solving for t:
0.5 * g * t^2 - viyt + y = 0
0.5 * (9.8) * t^2 - (10.14) * t + 29 = 0
Now, we can solve this quadratic equation for t using the quadratic formula.