Respuesta :
Answer:
a = 580 m/s^2
Explanation:
Given:
- Distance for accelerated throw s_a = 70 cm
- Angle of throw Q = 30 degrees
- Distance traveled by the javelin in horizontal direction x(f) = 75 m
- Initial height of throw y(0) = 0
- Final height of the javelin y(f) = -2 m
Find:
What was the acceleration of the javelin during the throw? Assume that it has a constant acceleration.
Solution:
- Compute initial components of the velocity:
V_x,i = V*cos(30)
V_y,i = V*sin(30)
- Use second equation of motion in horizontal direction:
x(f) = x(0) + V*cos(30)*t
75 = 0 + V*cos(30)*t
t = 75 /V*cos(30)
- Use equation of motion in vertical direction:
y(f) = y(0) + V_y,i*t + 0.5*g*t^2
Subs the values:
-2 = 0 + V*sin(30)*75/V*cos(30) - 4.905*(75/Vcos(30))^2
-2 = 75*tan(30) - 4.905*(5625/V^2*cos^2(30))
V^2 = 4.905*5625 / (2 + 75*tan(30))*cos^2(30)
V^2 = 812.0633
V = 28.5 m/s
- Use the third equation of motion in the interval of the throw:
V^2 = U^2 + 2*a*s_a
28.5^2 = 2*a*0.7
a = 580 m/s^2
The path of a projectile follows a parabolic trajectory;
The acceleration of the javelin during the throw is approximately 580.06 m/s²
The reason the value for the acceleration of the javelin is correct is as follows:
The known parameters;
The distance over which the javelin is accelerated = 70 cm = 0.7 m
Height of the javelin above the ground when it was released, y₀ = 2.0 m
Angle above the horizontal above which the javelin is thrown, θ = 30°
Distance away the javelin hits the ground, the range, R = 75 m
The required parameter;
The acceleration of the javelin during the throw
Solution:
The formula that gives the velocity with which the javelin was thrown, u, based on the range of a projectile, x, or R and the height when the javelin is thrown is the parabolic form of the projectile equation, presented here as follows;
[tex]y = \mathbf{y_0 + tan(\theta) \cdot x - \dfrac{g}{2 \cdot u^2 \cdot cos^2\theta} \cdot x^2}[/tex]
When x = R, then y = 0, we get;
[tex]0 = 2 + tan(30^{\circ}) \times 75 - \dfrac{9.81}{2 \cdot u^2 \cdot cos^230^{\circ}} \times 75^2[/tex]
Which gives;
[tex]0 = 2 + \sqrt{\dfrac{1}{3} } \times 75 - \dfrac{9.81}{2 \cdot u^2 \cdot \dfrac{3}{4} } \times 75^2[/tex]
[tex]2 + \sqrt{\dfrac{1}{3} } \times 75 = \dfrac{9.81}{2 \cdot u^2 \cdot \dfrac{3}{4} } \times 75^2[/tex]
[tex]u = \sqrt{ \dfrac{9.81}{2 \cdot\left(2 + \sqrt{\dfrac{1}{3} } \times 75\right) \cdot \dfrac{3}{4} } \times 75^2} \approx 28.497[/tex]
The velocity with which the javelin leaves the hand of the javelin thrower, u ≈ 28.497 m/s
The kinematic equation of motion for finding the acceleration is as follows;
u² = u₀² + 2·a·s
Where;
u = Velocity with which the javelin was thrown ≈ 28.497 m/s
u₀ = The speed of the javelin at the start = 0
s = Distance the javelin was accelerated before reaching release = 0.7 m
a = The acceleration of the javelin during the throw
Which gives;
u² = 0² + 2·a·s = 2·a·s
a = u²/(2·s)
a ≈ (28.497²)/(2 × 0.7) = 580.06
The acceleration of the javelin during the throw, a ≈ 580.06 m/s²
Learn more about the parabolic form of the projectile motion here:
https://brainly.com/question/20627626
