A bullet is fired straight up from a gun with a
muzzle velocity of 253 m/s.
Neglecting air resistance, what will be its
displacement after 2.9 s? The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of m
To find the displacement of the bullet after 2.9 seconds, we can use the equation of motion for vertical motion under constant acceleration.
The equation is:
d = v₀t + (1/2)at²
Where:
d = displacement
v₀ = initial velocity
t = time
a = acceleration due to gravity
Given:
v₀ = 253 m/s (muzzle velocity)
t = 2.9 seconds
a = -9.8 m/s² (acceleration due to gravity, negative because it acts in the opposite direction of motion)
Now, substitute the given values into the equation:
d = (253 m/s)(2.9 s) + (1/2)(-9.8 m/s²)(2.9 s)²
d = (253 m/s)(2.9 s) + (1/2)(-9.8 m/s²)(8.41 s²)
d = 732.7 m - 40.9709 m
d ≈ 691.7291 m
Therefore, the displacement of the bullet after 2.9 seconds is approximately 691.7291 meters.