A long jumper's take-off velocity is 10 m/s at an angle of 20o to the horizontal. His centre
of gravity is 1 m above the ground at take-off and is 0.5 m above the ground at landing.
a)How far does he displace his centre of gravity horizontally between take-off and landing?
b) Is this the optimal angle of take-off? If not, what would the approximate optimal angle be
(no calculation necessary)? Given that this velocity vector is typical of world class long
jumpers, why do they not increase their angle of take-off? [4]
(a) Compute the time T that is takes to raise the C.G. from 1 m above ground to maximum height and back to 0.5 m. The time going up is t1 = Vo sin 20/g = 0.3486 s. The CG rises 0.5961 m in that time. To reach a point 0.5 m lower (at landing) requires time
t2 = sqrt(2*1.0961/9.81) = 0.4727 s
Thus T = t1 + t2 = 0.8213 s
The horizontal travel of the CG is
10*cos20*T = 7.72 m
(b) No. The optimum is closer to 45 degrees at that takeoff velocity.
(c) Taking off at a steeper angl, such as 40-45 degrees would require a large sudden loss of velocity because, at the moment of the jump, the angular momentum about the takeoff foot cannot change, and the velocity must drop by a larger amount than it would with a 20 degree takeoff. That would have a bad effect on the length of the jump.
Notice what happens, for example, to the veloity of a pole vaulter after he or she plants the pole. Their takeoff speed immediately becomes less than the previous runniung speed. (The highly flexible poles now used make the effect less noticeable)
To find the displacement of the center of gravity horizontally between take-off and landing, we can use the horizontal component of the velocity vector. The horizontal component can be found using trigonometry.
a) The horizontal component of the velocity is calculated as follows:
Horizontal velocity = Initial velocity * cos(angle)
Given:
Initial velocity (Vi) = 10 m/s
Angle (θ) = 20°
Horizontal velocity = 10 m/s * cos(20°)
Horizontal velocity = 10 m/s * 0.9397
Horizontal velocity ≈ 9.397 m/s
Now, we can calculate the horizontal displacement by multiplying the horizontal velocity by the time of flight. To find the time of flight, we can use the equation:
Time of flight (t) = 2 * vertical component of velocity / acceleration due to gravity
The vertical component of velocity can be found using trigonometry:
Vertical velocity = Initial velocity * sin(angle)
Vertical velocity = 10 m/s * sin(20°)
Vertical velocity = 10 m/s * 0.3420
Vertical velocity ≈ 3.420 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Time of flight = 2 * 3.420 m/s / 9.8 m/s^2
Time of flight ≈ 0.6974 s
Now, we can calculate the horizontal displacement:
Horizontal displacement = Horizontal velocity * Time of flight
Horizontal displacement = 9.397 m/s * 0.6974 s
Horizontal displacement ≈ 6.5494 meters
Therefore, the center of gravity displaces approximately 6.5494 meters horizontally between take-off and landing.
b) The optimal angle of take-off for long jumpers is slightly less than 45 degrees. This angle maximizes the horizontal distance while still allowing a good vertical component to ensure a safe landing. If the angle of take-off is too high, the long jumper may not get enough horizontal distance. Similarly, if the angle of take-off is too low, the horizontal distance may be compromised. World-class long jumpers do not increase their angle of take-off beyond the optimal range because it would negatively affect their performance in terms of distance covered.
a) To find the horizontal displacement of the jumper's center of gravity between take-off and landing, we need to calculate the horizontal component of the velocity at take-off and landing.
Given:
Take-off velocity (v) = 10 m/s
Take-off angle (θ) = 20°
Height at take-off (h1) = 1 m
Height at landing (h2) = 0.5 m
First, we need to find the horizontal component of the velocity at take-off:
Horizontal component of velocity at take-off (v1x) = v * cos(θ)
v1x = 10 m/s * cos(20°) = 9.396 m/s
Next, we need to find the horizontal component of the velocity at landing:
To do this, we can use the conservation of mechanical energy principle, which states that the total mechanical energy (kinetic + potential) is conserved in the absence of external forces like air resistance. Since we are assuming no external forces in this scenario, we can use this principle to find the velocity at landing.
At take-off, the total mechanical energy is given by:
Total mechanical energy at take-off = 1/2 * m * v^2 + m * g * h1
At landing, the total mechanical energy is given by:
Total mechanical energy at landing = 1/2 * m * v2^2 + m * g * h2
Since the total mechanical energy is conserved, we can equate these two expressions:
1/2 * m * v^2 + m * g * h1 = 1/2 * m * v2^2 + m * g * h2
Simplifying and rearranging the equation:
v2^2 = v^2 + 2 * g * (h1 - h2)
Now we can substitute the given values:
v2^2 = (10 m/s)^2 + 2 * 9.8 m/s^2 * (1 m - 0.5 m)
v2^2 = 100 m^2/s^2 + 9.8 m/s^2 * 0.5 m
v2^2 = 100 m^2/s^2 + 4.9 m^2/s^2
v2^2 = 104.9 m^2/s^2
v2 ≈ 10.24 m/s (approximately)
The horizontal component of the velocity at landing (v2x) is then:
v2x ≈ v2 * cos(θ) ≈ 10.24 m/s * cos(20°) ≈ 9.61 m/s
Therefore, the horizontal displacement of the center of gravity (Δx) between take-off and landing is given by:
Δx = v2x - v1x
Δx = 9.61 m/s - 9.396 m/s
Δx ≈ 0.214 m
Therefore, the long jumper displaces his center of gravity approximately 0.214 meters horizontally between take-off and landing.
b) The approximate optimal angle of take-off for long jumpers is known to be around 45 degrees. This allows for the best trade-off between horizontal distance and vertical height achieved during the jump.
World-class long jumpers do not usually strive for higher take-off angles for a few reasons:
1. Horizontal distance: Increasing the take-off angle beyond 45 degrees would result in a greater vertical distance but would sacrifice horizontal distance. Since the objective of the long jump is to achieve the maximum horizontal distance, long jumpers aim to optimize the horizontal component of their velocity.
2. Technique and efficiency: The 45-degree angle provides a balance between jump height and jump distance, and world-class long jumpers have mastered the technique and timing required for this optimal angle. Deviating from this angle might require significant changes in technique and could reduce efficiency.
3. Risk of fouling: Long jumpers need to be mindful of not stepping over the take-off board, as this would result in a foul. A higher take-off angle increases the risk of stepping over, as the need for a longer horizontal distance becomes more critical.
Overall, while increasing the take-off angle might result in greater vertical height, it is not conducive to achieving the maximum horizontal distance that long jumpers strive for.