An arrow is shot straight up in the air with speed of 20m/s. How long is t in the air? How high will the arrow rise?

F = m a

F = -mg
so a = -g = -9.8
v = Vo -9.8t
0 at top = 20 - 9.8 t
t at top = 20/9.8 (same time falling so double that for time in air)
at t = t at top. how high is it?
h = 20 t - (1/2) (9.8) t^2

amm i got an answer of 4.08 for the time in the air I divided 20/9.8=2.04x2=4.08

For the rise I do not multiply by 2 t right? I got answer of 20.4

I am very glad that you were able to help me with this problem

To find the time the arrow is in the air, we can use the equation:

t = (2 * v) / g

where t is the time, v is the initial velocity, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the given values, we have:

t = (2 * 20 m/s) / 9.8 m/s²
t = 2.04 seconds

Therefore, the arrow is in the air for approximately 2.04 seconds.

To find the maximum height the arrow will reach, we can use the equation:

h = (v²) / (2 * g)

where h is the maximum height.

Substituting the given values, we have:

h = (20 m/s)² / (2 * 9.8 m/s²)
h = 204 / 19.6
h ≈ 10.41 meters

Therefore, the arrow will rise to a maximum height of approximately 10.41 meters.

To determine how long the arrow is in the air, we can use the fact that the initial vertical velocity when the arrow is shot up is 20 m/s, and the final vertical velocity when the arrow reaches the highest point is 0 m/s.

The time it takes for the arrow to reach its highest point can be found using the equation for vertical motion:

v(final) = v(initial) + (acceleration * t),

where v(final) is the final velocity, v(initial) is the initial velocity, acceleration is the acceleration due to gravity (approximated as 9.8 m/s²), and t is the time taken.

In this case, we have:

0 m/s = 20 m/s + (-9.8 m/s² * t).

By rearranging the equation, we can solve for t:

-20 m/s = -9.8 m/s² * t.

Dividing both sides of the equation by -9.8 m/s² yields:

t = 20 m/s / 9.8 m/s²,
t ≈ 2.04 seconds.

So, the arrow will be in the air for approximately 2.04 seconds.

To determine how high the arrow will rise, we can use the equation for vertical displacement during uniform acceleration:

s = v(initial) * t + (1/2) * acceleration * t²,

where s is the displacement (height), v(initial) is the initial velocity, acceleration is the acceleration due to gravity, and t is the time taken.

Plugging in the given values:

s = 20 m/s * 2.04 s + (1/2) * (-9.8 m/s²) * (2.04 s)²,

s ≈ 20.4 m.

Therefore, the arrow will rise to a height of approximately 20.4 meters.