An arrow is shot straight up into the air at 80.0 m/s.
a) What is the arrow’s maximum height?
b) How long does it take the arrow to reach its maximum height?
To solve these problems, we will use the equations of motion.
a) We can find the maximum height reached by the arrow using the equation:
v^2 = u^2 + 2as,
where v is the final velocity (0 m/s when the arrow reaches its maximum height), u is the initial velocity (80.0 m/s), a is the acceleration (which is -9.8 m/s^2 due to gravity), and s is the displacement.
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (0^2 - (80.0)^2) / (2 * -9.8)
s = -6400 / -19.6
s = 326.53 m
Therefore, the arrow's maximum height is 326.53 m.
b) To find the time it takes for the arrow to reach its maximum height, we can use the equation:
v = u + at
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values:
t = (0 - 80.0) / -9.8
t = 8.16 s
Therefore, it takes the arrow 8.16 seconds to reach its maximum height.
To answer these questions, we can use the equations of motion for uniformly accelerated motion.
Let's first find the time it takes for the arrow to reach its maximum height.
The initial velocity, u = 80.0 m/s (upwards)
The acceleration, a = -9.8 m/s^2 (gravity acting downwards)
Using the equation v = u + at, where v is the final velocity and t is the time:
0 = 80.0 - 9.8t
Rearranging the equation for t, we get:
t = 80.0 / 9.8
t ≈ 8.163 s (rounded to 3 decimal places)
So, it takes approximately 8.163 seconds for the arrow to reach its maximum height.
Now, let's find the maximum height reached by the arrow.
To determine the maximum height, we can use the equation:
h = u*t + (1/2) * a * t^2
Plugging in the values:
h = 80.0 * 8.163 + (1/2) * (-9.8) * (8.163)^2
h ≈ 332.04 m (rounded to 2 decimal places)
Therefore, the maximum height reached by the arrow is approximately 332.04 meters.
To find the answers to the questions, we can use kinematic equations of motion. The key equation we need is the equation for vertical displacement:
Δy = v0y * t + (0.5) * a * t^2
where:
Δy is the vertical displacement (maximum height in this case),
v0y is the initial vertical velocity (upward direction is positive),
t is the time,
and a is the acceleration due to gravity (-9.8 m/s^2 in the upward direction).
Now, let's proceed with the calculations.
a) What is the arrow’s maximum height?
To find the maximum height, we need to determine the time when the velocity of the arrow becomes zero as it reaches the peak of its trajectory. We can use the equation for vertical velocity:
v = v0y + a * t
When the arrow reaches its maximum height, v = 0. We can rearrange the equation to solve for t:
0 = v0y + (-9.8) * t
Solving for t:
t = -v0y / a
Substituting the given values:
v0y = 80.0 m/s
a = -9.8 m/s^2
t = -80.0 m/s / (-9.8 m/s^2)
t ≈ 8.16 s
Now that we have the time, we can find the maximum height using the equation for vertical displacement. We'll substitute the values:
Δy = v0y * t + (0.5) * a * t^2
Note that the acceleration due to gravity is -9.8 m/s^2 because it acts against the upward motion of the arrow.
Δy = (80.0 m/s) * (8.16 s) + (0.5) * (-9.8 m/s^2) * (8.16 s)^2
Δy ≈ 326.4 m
Therefore, the arrow's maximum height is approximately 326.4 meters.
b) How long does it take the arrow to reach its maximum height?
As calculated before, it takes approximately 8.16 seconds for the arrow to reach its maximum height.