You shoot an arrow into the air. Two seconds later (2.00 {\rm s}) the arrow has gone straight upward to a height of 32.0 m above its launch point.
What is the initial velocity and how long did it take for the arrow to reach 16.0 m above the launch point?
Velocity (v) obeys the equation
v = Vo - gt = Vo - 9.8 t
where Vo is the initial velocity
At t=2 s
32 = Vo - 19.6 m
Therefore Vo = 51.6 m/s
y (the height) = Vo t - (g/2)t^2
= 51.6 t - 4.9 t^2
Set y = 16 and solve for the time t when that height is reached. Take the lowest of the two solutions of the quadratic equation; the other is the time coming back down.
To calculate the initial velocity and the time taken for the arrow to reach a certain height, we can use the equations of motion for projectile motion. In this case, because the arrow is shot straight upward, we only need to consider the vertical motion.
Let's denote the following variables:
- Initial velocity (u)
- Final velocity (v) = 0 (because the arrow reaches its highest point)
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)
- Displacement (s) = 32.0 m (highest point reached by the arrow)
Using the equation of motion for displacement (s), initial velocity (u), and time (t):
s = ut + (1/2)gt^2
Plugging in the given values, we have:
32.0 = u(2.00) + (1/2)(9.8)(2.00)^2
Simplifying the equation, we get:
32.0 = 2u + 19.6
Rearranging the equation to solve for u:
2u = 32.0 - 19.6
2u = 12.4
u = 6.2 m/s
So, the initial velocity of the arrow is 6.2 m/s.
Now, to calculate the time it takes for the arrow to reach 16.0 m above the launch point, we can use the same equation of motion and solve for time (t).
Using the given values:
s = ut + (1/2)gt^2
16.0 = 6.2t + (1/2)(9.8)t^2
Simplifying the equation, we get:
16.0 = 6.2t + 4.9t^2
Rearranging the equation to obtain a quadratic equation:
4.9t^2 + 6.2t - 16.0 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Using the values a = 4.9, b = 6.2, and c = -16.0, we can substitute them into the quadratic formula and solve for t:
t = (-6.2 ± sqrt(6.2^2 - 4(4.9)(-16.0))) / (2(4.9))
Calculating the values inside the square root:
t = (-6.2 ± sqrt(38.44 + 313.6)) / 9.8
t = (-6.2 ± sqrt(352.04)) / 9.8
t = (-6.2 ± 18.76) / 9.8
Using both the positive and negative solutions, we have two possible times. However, since we are only interested in the time it takes for the arrow to reach the 16.0 m mark, we can disregard the negative value.
Taking the positive value:
t = (-6.2 + 18.76) / 9.8
t = 12.56 / 9.8
t ≈ 1.28 s
So, it took approximately 1.28 seconds for the arrow to reach 16.0 m above the launch point.