It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?
Please help! I do not have any clue as to how to solve. Thanks!
I don't quite understand how you determined the time without it being given to you.
Thanks
It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?
The general equations of accelerated motion apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g). This results in
....Vf = Vo + gt (the acceleration is assumed constant)
....d = Vo(t) + g(t^2)/2
....Vf^2 = Vo^2 + 2gd
As written, these expressions apply to falling bodies. The equations that apply to rising bodies are
....Vf = Vo - gt (the acceleration is assumed constant)
....d = Vo(t) - gt^2/2
....Vf^2 = Vo^2 - 2gd
From Vf = 0 = 100 - 9.8t, t(up) = 10.2sec. = t(dwn) making the total flight time 20.4sec.
From h = Vot - gt^2/2,
h = 100(10.2) - 9.8(10.2)^2/2 or
h = 510.2m.
To solve these questions, we can use the equations of motion to analyze the arrow's vertical motion. Let's tackle each question one by one:
(a) To determine how high the arrow would rise, we need to find the maximum height reached by the arrow. At the highest point, the arrow's vertical velocity will be zero. We can use the equation for vertical motion:
v^2 = u^2 + 2as
Where:
- v is the final velocity (zero at maximum height)
- u is the initial velocity (100 m/s)
- a is acceleration due to gravity (-9.8 m/s^2, assuming we are on Earth)
- s is displacement (maximum height reached)
Simplifying the equation for our purpose, we get:
0 = (100 m/s)^2 + 2(-9.8 m/s^2)s
Solving for s, we have:
-9800s = -10000
s = -10000 / -9800
s ≈ 1.02 meters
Therefore, neglecting the effects of friction, the arrow would rise to a height of approximately 1.02 meters.
(b) To determine how long the arrow would be in the air, we need to find the time it takes to reach the maximum height and then double it, as the total time in the air will be the time it takes to reach the maximum height plus the time it takes to fall back down to the ground.
To find the time it takes to reach the maximum height, we can use the equation:
v = u + at
Where:
- v is the final velocity (zero at maximum height)
- u is the initial velocity (100 m/s)
- a is acceleration due to gravity (-9.8 m/s^2)
- t is time
Substituting the given values, we have:
0 = 100 m/s + (-9.8 m/s^2)t
Simplifying the equation for t, we have:
-9.8t = -100
t = -100 / -9.8
t ≈ 10.2 seconds
Since the maximum height is reached halfway through the total time in the air, the time for the arrow to reach the maximum height is half of the total time. Therefore, the total time in the air is approximately:
2 * 10.2 seconds = 20.4 seconds
Hence, neglecting the effects of friction, the arrow would be in the air for approximately 20.4 seconds.