A bullet is fired straight up from a gun with a

muzzle velocity of 225 m/s.
Neglecting air resistance, what will be its
displacement after 6.1 s? The acceleration of
gravity is 9.8 m/s2 .
Answer in units of m

To find the displacement of the bullet after 6.1 s, we can use the kinematic equation for displacement:

š‘‘ = š‘£ā‚€š‘” + Ā½š‘Žš‘”Ā²

Where:
š‘‘ is the displacement,
š‘£ā‚€ is the initial velocity,
š‘Ž is the acceleration, and
š‘” is the time.

In this case, the bullet is fired straight up, so the initial velocity is 225 m/s. The acceleration is the acceleration due to gravity, which is -9.8 m/sĀ² (negative because it acts in the opposite direction to the velocity).

Substituting the values into the equation:

š‘‘ = (225 m/s) * (6.1 s) + Ā½ * (-9.8 m/sĀ²) * (6.1 s)Ā²

Simplifying the equation gives:

š‘‘ = (225 m/s) * (6.1 s) - Ā½ * (9.8 m/sĀ²) * (6.1 s)Ā²

š‘‘ = 1372.5 m - 179.263 m

š‘‘ ā‰ˆ 1193.24 m (rounded to two decimal places)

Therefore, the displacement of the bullet after 6.1 s, neglecting air resistance, is approximately 1193.24 m.

To find the displacement of the bullet after 6.1 seconds, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = displacement
u = initial velocity
t = time
a = acceleration due to gravity

Given:
u = 225 m/s (muzzle velocity)
t = 6.1 s
a = 9.8 m/s^2 (acceleration due to gravity)

Substituting the given values into the equation, we have:

s = (225 * 6.1) + (0.5 * 9.8 * 6.1^2)

s = 1372.5 + 0.5 * 9.8 * 37.21

s = 1372.5 + 181.819

s ā‰ˆ 1554.319

Therefore, the displacement of the bullet after 6.1 seconds is approximately 1554.319 meters.