An archer fires an arrow straight up into the air with a speed vo = 14 m/s. Neglect air resistance.
Find the maximum height h reached by the arrow, in meters.
Write an expression for the time the arrow is in the air until it returns to launch height in terms of known quantities.
K.E. at launch equals P.E. at peak
1/2 m v^2 = m g h ... h = v^2 / (2 g)
0 = -4.9 t^2 + 14 t
what is the expression for time?
0 = -4.9 t^2 + 14 t = t (14 - 4.9 t)
so h = 0 at t = 0 and at t = 14 / 4.9
To find the maximum height reached by the arrow, we can use the kinematic equation for vertical motion.
The equation is: vf^2 = vo^2 + 2gh
where vf is the final velocity (which is zero at maximum height), vo is the initial velocity, g is the acceleration due to gravity, and h is the maximum height.
Since the arrow is launched vertically upward, the initial and final velocities are opposite in direction. Thus, we have:
0 = (14 m/s)^2 + 2(-9.8 m/s^2)h
Simplifying the equation gives us:
196 = 19.6h
Dividing both sides by 19.6, we find:
h = 10 m
Therefore, the maximum height reached by the arrow is 10 meters.
To write an expression for the time the arrow is in the air until it returns to launch height, we need to consider the time it takes for the arrow to reach the maximum height and then return to the launch height.
The time to reach the maximum height can be found using the equation:
vf = vo + gt
Since the final velocity at the maximum height is zero, we have:
0 = 14 m/s + (-9.8 m/s^2)t
Solving for t gives us:
t = 14 m/s / 9.8 m/s^2 = 1.43 s
The total time in the air can be found by doubling the time to reach the maximum height:
Total time = 2 * 1.43 s = 2.86 s
Therefore, the expression for the time the arrow is in the air until it returns to launch height is:
t_total = 2 * (vo / g)