A bullet is fired straight up from a gun with a
muzzle velocity of 259 m/s.
Neglecting air resistance, what will be its displacement after 8.9 s? The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of m. Please show work
h=vi*t-4.9t^2
Thanks it was a huge help.
To find the displacement of the bullet after 8.9 seconds, we can use the equation of motion:
s = ut + (1/2)at²
Where:
s = displacement
u = initial velocity
t = time
a = acceleration
Given:
u = 259 m/s (upward direction)
t = 8.9 s
a = -9.8 m/s² (negative because gravity is acting downward)
Substituting the values into the equation:
s = (259 m/s)(8.9 s) + (1/2)(-9.8 m/s²)(8.9 s)²
First, calculate the first term:
259 m/s * 8.9 s = 2,301.1 m
Next, calculate the second term:
(1/2)(-9.8 m/s²)(8.9 s)² = -350.476 m
Add the two terms together:
2,301.1 m + (-350.476 m) = 1,950.624 m
Therefore, the displacement of the bullet after 8.9 seconds is approximately 1,950.624 meters.