An athlete performing a long jump leaves the ground at a 27.1 ∘ angle and lands 7.73 m away.
1). What was the takeoff speed?
Express your answer using three significant figures and include the appropriate units.
2).If this speed were increased by just 3.0%, how much longer would the jump be?
Express your answer using two significant figures and include the appropriate units.
the distance equals the flight time multiplied by the horizontal component of the takeoff speed
t = 2 {[s * sin(27.1º)] / g}
d = t * [s * cos(27.1º)]
7.73 = 2 * s^2 * sin(27.1º / 2) / g
To solve these problems, we can use the kinematic equations of motion. Let's start with the first question:
1) To calculate the takeoff speed, we need to determine the initial horizontal and vertical components of the velocity.
Given:
Angle of takeoff (θ) = 27.1°
Distance traveled (d) = 7.73 m
In the long jump, the horizontal and vertical components of velocity are independent of each other.
The horizontal component of velocity (Vx) remains constant throughout the motion, while the vertical component of velocity (Vy) changes due to the effect of gravity.
By breaking down the initial velocity (V) into its horizontal (Vx) and vertical (Vy) components, we have:
Vx = V * cos(θ)
Vy = V * sin(θ)
The distance traveled horizontally (d) can be calculated by the formula:
d = Vx * t
Since there is no vertical displacement, we can use the vertical component's equation:
0 = Vy * t - 1/2 * g * t^2
Here, g represents the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.
Using the equations above, we can solve for V and t. First, calculate the time of flight (t) from the vertical component's equation:
0 = Vy * t - 1/2 * g * t^2
0 = V * sin(θ) * t - 1/2 * g * t^2
Next, solve for t:
t * (V * sin(θ) - 1/2 * g * t) = 0
Since t ≠ 0, we have:
V * sin(θ) - 1/2 * g * t = 0
Solving for t:
t = (2 * V * sin(θ)) / g
Now, substitute this value of t into the horizontal component's equation:
d = Vx * t
d = V * cos(θ) * [(2 * V * sin(θ)) / g]
Substituting the given values:
7.73 m = V * cos(27.1°) * [(2 * V * sin(27.1°)) / 9.8 m/s^2]
Simplifying the equation, we get:
V^2 = (7.73 m * 9.8 m/s^2) / (cos(27.1°) * sin(27.1°))
Taking the square root of both sides:
V = √[(7.73 m * 9.8 m/s^2) / (cos(27.1°) * sin(27.1°))]
Calculating this value, we find:
V ≈ 12.0 m/s
Therefore, the takeoff speed is approximately 12.0 m/s.
Now let's proceed to the second question:
2) If the speed were increased by just 3.0%, we can calculate the new distance traveled (d') using the relation between speed and distance.
If the speed is increased by 3.0%, the new speed (V') is given by:
V' = V + (0.03 * V)
Substituting the value of V we calculated earlier:
V' = 12.0 m/s + (0.03 * 12.0 m/s)
Simplifying, we get:
V' ≈ 12.4 m/s
To find the new distance traveled (d'), we can use the same formula as before:
d' = V' * cos(θ) * [(2 * V' * sin(θ)) / g]
Substituting the given values:
d' = (12.4 m/s) * cos(27.1°) * [(2 * 12.4 m/s * sin(27.1°)) / 9.8 m/s^2]
Calculating this value, we find:
d' ≈ 8.8 m
Therefore, if the speed is increased by just 3.0%, the jump would be approximately 8.8 meters long, which is 1.07 meters longer than the original jump.
Note: We used approximations during the calculations, so the final answers may have slight variations based on these approximations.