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
