Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is 34.7 m/s and that the second arrow is fired 1.55 s after the first. Determine the initial speed of the second arrow.
First Arrow.
Tr = (Vf-Vo)/g = (0-34.7) / -9.8=3.54s.
+ Riswe time.
hmax = (Vf^2-Vo^2)/2g.
hmax = (0-1204.1) / -19.6 = 61.4 m.
2nd Arrow.
Vf = Vo + gt.
Vo = Vf - gt
Vo = 0 - (-9.8)(3.54-1.55) = 19.5 m/s.
To determine the initial speed of the second arrow, we can use the principles of projectile motion.
Let's analyze the motion of the first arrow:
1. The initial velocity of the first arrow, u1, is given as 34.7 m/s.
2. The acceleration due to gravity, g, is approximately 9.8 m/s^2 and acts downwards.
3. The time taken by the first arrow to reach its maximum height is unknown, let's call it t_max1.
4. At the maximum height, the vertical component of velocity becomes zero.
5. The final vertical velocity of the first arrow at maximum height, v1_max, is unknown.
Using the second and fourth principles of projectile motion, we can write an equation for the vertical component of velocity at the maximum height:
v1_max = u1 - g * t_max1 (Equation 1)
Now let's analyze the motion of the second arrow:
1. The initial velocity of the second arrow, u2, is unknown.
2. The time taken by the second arrow to reach its maximum height is given as 1.55 s.
3. The acceleration due to gravity, g, is approximately 9.8 m/s^2 and acts downwards.
4. At the maximum height, the vertical component of velocity becomes zero.
5. The final vertical velocity of the second arrow at maximum height, v2_max, is unknown.
Using the second and fourth principles of projectile motion, we can write an equation for the vertical component of velocity at the maximum height:
v2_max = u2 - g * t_max2 (Equation 2)
Since both arrows reach their maximum heights at the same instant, we can equate t_max1 and t_max2:
t_max1 = t_max2 + 1.55 s
Now, since the vertical velocity at maximum height is zero for both arrows, we can set v1_max equal to v2_max:
v1_max = v2_max
Substituting the equations (Equation 1 and Equation 2) and rearranging, we get:
u1 - g * t_max1 = u2 - g * t_max2
We can substitute the value of t_max1 from the equation t_max1 = t_max2 + 1.55 s, and rearrange the equation to solve for u2:
u2 = u1 - g * 1.55 s
Substituting the given values, we have:
u2 = 34.7 m/s - 9.8 m/s^2 * 1.55 s
u2 = 34.7 m/s - 15.19 m/s
u2 = 19.51 m/s
Therefore, the initial speed of the second arrow is approximately 19.51 m/s.