A soccer ball iskicked with an intial horizontal speed of 6 m/s and initial vertical speed of 3.5 m/s. Assume that the projection and landing height are the same, and neglect air resistance. Calculate the following
a) the ball's projection speed and angle
b)the ball's horizontal speed at .5 s into its flight
c) The ball's vertical speed at .5 s into its flight
d) the ball's height at its apex
e) the ball's flight time
f)the ball's flight distance
Xo = 6 m/s.
Yo = 3.5 m/s.
a. tanA = Yo/Xo = 3.5/6 = 0.58333
A = 30.3o
Vo=Xo/cosA = 6/cos30.3 = 6.95m/s[30.3o]
b. X = Xo = 6 m/s. It does not change.
c. Y = Yo + g*t
Y = 3.5 - 9.8*0.5 = -1.4 m/s. The negative sign means the ball is falling.
d. h=(Y^2-Yo^2)/2g=(0-3.5^2)/-19.6 = 0.625 m.
e. Y = Yo + g*t = 0 @ max. ht.
Tr = (V-Vo)/g = (0-3.5)/-9.8=0.357 s =
Rise time.
Tf = Tr = 0.357 s. = Fall time.
T = Tr + Tf = 0.357 + 0.367 = 0.714 s.
f. D = Xo * t = 6m/s * 0.714s = 4.29 m.
= 0.357 = 0.714 s.
a) To find the ball's projection speed and angle, we can use trigonometry. The projection speed (also known as the initial velocity) is the magnitude of the vector formed by the ball's initial horizontal and vertical speeds. We can calculate it using the Pythagorean theorem:
Initial speed = √(horizontal speed^2 + vertical speed^2)
= √(6 m/s)^2 + (3.5 m/s)^2)
= √(36 m^2/s^2 + 12.25 m^2/s^2)
= √(48.25 m^2/s^2)
≈ 6.95 m/s
To find the angle, we can use inverse tangent:
Angle = arctan(vertical speed / horizontal speed)
= arctan(3.5 m/s / 6 m/s)
≈ 30.96 degrees
Therefore, the ball's projection speed is approximately 6.95 m/s, and the angle is approximately 30.96 degrees.
b) To find the ball's horizontal speed at 0.5 s into its flight, we need to know the horizontal acceleration. Since there is no horizontal acceleration, the horizontal speed remains constant throughout the motion. Therefore, the horizontal speed at any given time remains 6 m/s.
c) The ball's vertical speed at 0.5 s into its flight can be found using the equation for uniform acceleration:
Final vertical speed = Initial vertical speed + (acceleration * time)
Since the ball's initial vertical speed is 3.5 m/s and there is no vertical acceleration, the vertical speed remains constant throughout the motion. Therefore, the vertical speed at any given time remains 3.5 m/s.
d) The height at the apex (highest point) of the ball's trajectory can be found using the kinematic equation for vertical motion. At the apex, the ball's vertical speed becomes zero.
Final vertical speed = Initial vertical speed + (acceleration * time)
0 = 3.5 m/s + (-9.8 m/s^2 * time)
Solving for time:
Time = -3.5 m/s / (-9.8 m/s^2)
≈ 0.36 seconds
Substituting the time back into the equation:
Height at apex = Initial vertical speed * time + (0.5 * acceleration * time^2)
= 3.5 m/s * 0.36 s + (0.5 * -9.8 m/s^2 * (0.36 s)^2)
≈ 0.63 meters
Therefore, the ball's height at its apex is approximately 0.63 meters.
e) The ball's flight time is the total time it takes for the ball to reach its starting height after being kicked.
To find the flight time, we can double the time it takes to reach the apex. From previous calculations, we found that it takes about 0.36 seconds to reach the apex. Therefore, the total flight time is approximately 0.36 s * 2 = 0.72 seconds.
f) The ball's flight distance can be calculated using the horizontal speed and the flight time.
Flight distance = Horizontal speed * Flight time
= 6 m/s * 0.72 s
≈ 4.32 meters
Therefore, the ball's flight distance is approximately 4.32 meters.