An arrow is shot straight upward at 55 m/s.
a) find the time until the arrow reaches its peak
b)how high was the arrow at its peak
c)find the total time the arrow is in the air
d)what was the velocity of the arrow after 3.5 seconds
e)how high was it at 3.5 seconds
To answer these questions, we can use the equations of motion for linear projectile motion. These equations will help us determine the time, height, and velocity at different points during the arrow's flight.
1. To find the time until the arrow reaches its peak:
The initial vertical velocity of the arrow is 55 m/s (positive, since it goes upward). At the highest point, the vertical velocity will become zero. We can use the following equation to solve for the time it takes to reach this point:
vf = vi + gt
where:
vf = final velocity (0 m/s at the peak)
vi = initial velocity (55 m/s)
g = acceleration due to gravity (-9.8 m/s², since it acts in the opposite direction of the initial velocity)
Setting vf = 0 and solving for time (t), we get:
0 = 55 - 9.8t
9.8t = 55
t = 55 / 9.8
Using a calculator, we find that t ≈ 5.61 seconds.
Therefore, it takes approximately 5.61 seconds for the arrow to reach its peak.
2. To find the height of the arrow at its peak:
The height at the peak can be determined using the following equation:
h = vi(t) + (1/2)gt²
where:
h = height at the peak
vi = initial velocity (55 m/s)
t = time it takes to reach the peak (5.61 seconds)
g = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we can calculate:
h = 55(5.61) + (1/2)(-9.8)(5.61)²
Using a calculator, we find that h ≈ 153.28 meters.
Therefore, the arrow reaches a height of approximately 153.28 meters at its peak.
3. To find the total time the arrow is in the air:
The total time the arrow is in the air is twice the time it takes to reach the peak, as it will take the same amount of time to fall back down. Therefore, the total time is:
Total time = 2 * t
Total time = 2 * 5.61
Using a calculator, we find that the total time in the air is approximately 11.22 seconds.
4. To find the velocity of the arrow after 3.5 seconds:
We can use the equation for instantaneous velocity:
v = vi + gt
where:
v = final velocity after t seconds
vi = initial velocity (55 m/s)
g = acceleration due to gravity (-9.8 m/s²)
t = time (3.5 seconds)
Plugging in the values, we can calculate:
v = 55 + (-9.8)(3.5)
Using a calculator, we find that v ≈ 20.3 m/s.
Therefore, the velocity of the arrow after 3.5 seconds is approximately 20.3 m/s.
5. To find the height of the arrow at 3.5 seconds:
To find the height at any particular time during its flight, we can use the equation:
h = vi(t) + (1/2)gt²
where:
h = height at the specific time
vi = initial velocity (55 m/s)
t = time (3.5 seconds)
g = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we can calculate:
h = 55(3.5) + (1/2)(-9.8)(3.5)²
Using a calculator, we find that h ≈ 87.64 meters.
Therefore, the height of the arrow at 3.5 seconds is approximately 87.64 meters.