A ball is kicked with an initial velocity of
20.0 m/s
at an angle of
35 degrees
to the ground. If
it is to
pass over a wall 5.6 m high determine
a.
The minimum horizontal the wall can be placed from the ball’s launch point in
meters.
b.
What would be the ball’s velocity just as it goes over the wall in m/s?
c.
Provid
ed it clears the wall, what is the range of the ball in meters?
To solve these problems, we can break down the initial velocity of the ball into its horizontal and vertical components.
Given:
Initial velocity (v₀) = 20.0 m/s
Angle (θ) = 35 degrees
Height of the wall (h) = 5.6 m
a. To determine the minimum horizontal distance the wall can be placed from the ball's launch point, we need to find the time it takes for the ball to reach the height of the wall.
The vertical component of velocity (vₓ) can be calculated using the formula:
vₓ = v₀ * cos(θ)
Plugging in the values:
vₓ = 20.0 * cos(35) ≈ 16.4 m/s
To find the time it takes for the ball to reach the height of the wall, we use the equation for vertical motion:
h = v₀y * t + (1/2) * g * t²
Where:
v₀y is the initial vertical component of velocity,
t is the time, and
g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the ball is launched vertically, the initial vertical component of velocity (v₀y) is zero.
Substituting the values:
h = 0 + (1/2) * (9.8) * t² (because v₀y = 0)
Simplifying:
5.6 = 4.9 * t²
Rearranging the equation:
t² = 5.6 / 4.9
t² ≈ 1.14
t ≈ √(1.14)
t ≈ 1.07 s
To find the horizontal distance (d) traveled by the ball, we use the equation:
d = vₓ * t
Substituting the values:
d = 16.4 * 1.07
d ≈ 17.50 m
Therefore, the minimum horizontal distance the wall can be placed from the ball's launch point is approximately 17.50 meters.
b. To find the ball's velocity just as it goes over the wall, we need to find the vertical component of velocity (v_y).
The vertical component of velocity (v_y) can be calculated using the formula:
v_y = v₀ * sin(θ)
Plugging in the values:
v_y = 20.0 * sin(35) ≈ 11.47 m/s
The velocity just as it goes over the wall is the resultant velocity, which can be found using the Pythagorean theorem:
v = √(vₓ² + v_y²)
Plugging in the values:
v = √(16.4² + 11.47²)
v ≈ √(268.96 + 131.8609)
v ≈ √(400.8209)
v ≈ 20.02 m/s
Therefore, the ball's velocity just as it goes over the wall is approximately 20.02 m/s.
c. To find the range of the ball, we need to find the horizontal distance traveled by the ball.
The range (R) can be calculated using the formula:
R = (v₀² * sin(2θ)) / g
Plugging in the values:
R = (20.0² * sin(2 * 35)) / 9.8
R = (400 * sin(70)) / 9.8
R ≈ (400 * 0.9397) / 9.8
R ≈ 37.99 m
Therefore, the range of the ball, provided it clears the wall, is approximately 37.99 meters.
To determine the answers to these questions, we can use the equations of projectile motion. Projectile motion occurs when an object is launched into the air and moves along a curved path due to the influence of gravity. The key variables we need to consider are the initial velocity, launch angle, and height of the wall.
a. To find the minimum horizontal distance the wall can be placed from the ball's launch point, we need to calculate the range. The range is the horizontal distance traveled by the projectile before it hits the ground. The formula for the range (R) in projectile motion is:
R = (v^2 * sin(2θ)) / g
where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values:
v = 20.0 m/s
θ = 35 degrees
g = 9.8 m/s^2
R = (20.0^2 * sin(2 * 35)) / 9.8
R = (400 * sin(70)) / 9.8
R ≈ 167.19 m
Therefore, the minimum horizontal distance the wall can be placed from the ball's launch point is approximately 167.19 meters.
b. To find the ball's velocity just as it goes over the wall, we can analyze the vertical and horizontal components separately. The vertical component of velocity remains the same throughout the motion, but the horizontal component affects the range.
The vertical component of velocity (Vy) can be found using the formula:
Vy = v * sin(θ)
Substituting the given values:
v = 20.0 m/s
θ = 35 degrees
Vy = 20.0 * sin(35)
Vy ≈ 11.42 m/s
The horizontal component of velocity (Vx) can be found using the formula:
Vx = v * cos(θ)
Substituting the given values:
v = 20.0 m/s
θ = 35 degrees
Vx = 20.0 * cos(35)
Vx ≈ 16.41 m/s
Since the ball clears the wall, at the highest point of its trajectory, its velocity will only have a vertical component. Therefore, the ball's velocity just as it goes over the wall is approximately 11.42 m/s.
c. To find the range of the ball, we can use the same formula as in part a:
R = (v^2 * sin(2θ)) / g
Substituting the given values:
v = 20.0 m/s
θ = 35 degrees
g = 9.8 m/s^2
R = (20.0^2 * sin(2 * 35)) / 9.8
R = (400 * sin(70)) / 9.8
R ≈ 167.19 m
Therefore, the range of the ball, provided it clears the wall, is approximately 167.19 meters.