a soccer player takes a penalty kick 11.0m from the goal. to score she has to shoot the ball under the 2.44m high crossbar. if she kicks at 19.8m/s, what range of launch angles will result in a goal?
To find the range of launch angles that will result in a goal, we need to consider the height of the crossbar and the initial velocity of the ball. The range of angles can be determined by considering the projectile motion of the ball.
Let's break down the problem into the horizontal and vertical components of motion:
1. Horizontal component: When a projectile is launched at an angle, the horizontal component of its velocity remains constant throughout its motion. In this case, the horizontal component is the same as the initial velocity of the ball, which is 19.8 m/s.
2. Vertical component: The vertical component of the ball's initial velocity will determine the height it reaches. We need to find the launch angles that allow the ball to clear the crossbar, which is 2.44 m high.
To find the range of launch angles, we can use the equations of projectile motion. The height reached by a projectile depends on its initial vertical velocity and the time it takes to reach the peak of its trajectory.
1. The formula for the maximum height (H) reached by a projectile is given by:
H = (V₀y²) / (2g)
Where V₀y is the initial vertical velocity and g is the acceleration due to gravity (approximately 9.8 m/s²).
2. We can rearrange the formula to solve for V₀y:
V₀y = √(2gH)
3. For the ball to clear the crossbar, the maximum height (H) should be greater than the height of the crossbar, which is 2.44 m.
√(2gH) > 2.44
4. Substituting the known values:
√(2 * 9.8 * H) > 2.44
Now that we have the equation, we can solve it for H to find the range of launch angles that will result in a goal.
To find the range of launch angles that will result in a goal, we need to consider the maximum height the ball can reach and the distance it needs to travel to reach the goal.
Given:
Initial velocity (u) = 19.8 m/s
Distance (d) = 11.0 m
Height of the crossbar (h) = 2.44 m
1. Let's find the time of flight (t) of the ball.
Using the equation: d = ut + (1/2)at²,
where a is the acceleration due to gravity (approximately -9.8 m/s²), since it acts in the opposite direction to the ball's initial velocity.
Rearranging the equation to find time:
t = (d - u * t) / (-4.9)
Substituting the values:
t = (11.0 - 19.8 * t) / (-4.9)
Simplifying:
4.9t = 11.0 - 19.8t
24.7t = 11.0
t ≈ 0.445 seconds
2. Now, let's find the maximum height (H) the ball will reach.
Using the equation: v² = u² + 2aH,
where v is the final velocity (0 m/s) at the maximum height.
Substituting the values:
0² = (19.8)² + 2 * (-9.8) * H
0 = 392.04 - 19.6H
19.6H = 392.04
H ≈ 20.01 m
3. Since the crossbar is at a height of 2.44 m, the ball needs to clear a height (h) of H - h to result in a goal.
20.01 - h ≥ 2.44
Simplifying:
h ≤ 20.01 - 2.44
h ≤ 17.57 m
4. Finally, let's find the range of launch angles (θ) using the equation: tan(θ) = (H - h) / (d/2)
Substituting the values:
tan(θ) = (20.01 - 2.44) / (11.0/2)
tan(θ) ≈ 1.505
Taking the inverse tangent (arctan) of both sides:
θ ≈ arctan(1.505)
θ ≈ 55.04°
So, the range of launch angles that will result in a goal is approximately from 0° to 55.04°.