An archer shoots an arrow that leaves the bow at an angle of 42 degrees above the horizontal. The arrow hits a target located 135 m away and 23 m above the height from which the arrow was shot. Given that the mass of the arrow is .125 kg and that it remained in contact with the string 1.28 m, what was the force applied to the arrow by the string of the bow?
Can someone at least tell me where to begin? I'm lost when it comes to these types of problems.
vertical direction equations:
v = Vo sin 42 -9.8 t
or
v = .669 Vo - 9.8 t
h = Ho + .669 Vo t - 4.9 t^2
Horizontal direction equation
x = (Vo cos 42 ) t
or
x = .743 Vo t
Now horizontal fact
135 = .743 Vo t
And vertical
23 = 0 + ,669 Vo t - 4.9 t^2
from horizontal equation
Vo t = 182
use that in vertical equation
23 = .669 (182) - 4.9 t^2
4.9 t^2 = 98.8
t = 4.49 seconds in air
so Vo = 182/4.49
Vo = 40.5 m/s
Now impuse = change of momentum
F d = m Vo
F (1.28) = .125 (40.5)
solve for F
To solve this problem, you can use the principle of conservation of mechanical energy, along with some basic trigonometry and kinematics equations. Here's a step-by-step guide on how to approach this problem:
1. Break down the problem into two separate parts: the horizontal motion and the vertical motion of the arrow.
2. Start by analyzing the horizontal motion. Since there are no horizontal forces acting on the arrow after it leaves the bow, the only force acting on the arrow is the initial force applied by the bowstring. This force will cause the arrow to move horizontally at a constant velocity.
3. Use the horizontal motion to find the initial velocity of the arrow. This can be done by using the equation v = d/t, where v is the velocity, d is the horizontal distance (135 m), and t is the time it takes for the arrow to travel that distance.
4. Now, let's focus on the vertical motion of the arrow. We know that the arrow was shot at an angle of 42 degrees above the horizontal. This means that the initial velocity of the arrow can be split into its horizontal and vertical components using trigonometry.
5. Use trigonometry to find the initial vertical velocity (v_y) and the initial horizontal velocity (v_x) of the arrow. The vertical component can be found using v_y = v * sin(42°), and the horizontal component can be found using v_x = v * cos(42°), where v is the initial velocity.
6. Since the arrow remained in contact with the string for 1.28 m, you can use the equation v = u + at, where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration, and t is the time taken. Rearrange the equation to solve for t.
7. Now that you know the time it took for the arrow to reach the target height, use the equation s = ut + (1/2)at^2 to find the vertical displacement (s) of the arrow.
8. To find the acceleration (a), consider that the only vertical force acting on the arrow is the force due to gravity. Therefore, you can use the equation a = g = 9.8 m/s^2, where g is the acceleration due to gravity.
9. Use the equation s = ut + (1/2)at^2, where s is the vertical displacement, u is the initial velocity, and a is the acceleration, to find the initial vertical velocity (u).
10. Finally, use the equation F = ma to find the force applied to the arrow by the string of the bow. Plug in the mass of the arrow (.125 kg) and the initial vertical acceleration (a) to calculate the force (F).
By following these steps, you should be able to solve the problem and find the force applied to the arrow by the bowstring.