A bullet is fired straight up from a gun with a
muzzle velocity of 170 m/s.
Neglecting air resistance, what will be its
displacement after 5.8 s? The acceleration of
gravity is 9.8 m/s2 .
Answer in units of m.
The vertical distance the bullet rises is
Y = 170*t - (g/2) t^2
Plug in g = 9.8 m/s^2 and t = 5.8 s and see what you get.
To calculate the displacement of the bullet after 5.8 seconds, we need to use the kinematic equation:
s = ut + (1/2)at^2
where:
- s is the displacement
- u is the initial velocity
- t is the time
- a is the acceleration
In this case, the bullet is fired straight up, so its initial velocity is positive 170 m/s. The acceleration due to gravity is downward, so it has a value of -9.8 m/s^2.
Substituting these values into the equation:
s = (170 m/s)(5.8 s) + (1/2)(-9.8 m/s^2)(5.8 s)^2
Simplifying:
s = 986 m - 161.672 m
s = 824.328 m
Therefore, the displacement of the bullet after 5.8 seconds is approximately 824.328 meters.
To find the displacement of the bullet after 5.8 seconds, we need to use the equation of motion for an object in free fall. The equation is:
s = ut + (1/2)gt^2
Where:
s is the displacement
u is the initial velocity
t is the time
g is the acceleration due to gravity
In this case, the initial velocity of the bullet is the muzzle velocity, u = 170 m/s. The time, t = 5.8 s. And the acceleration due to gravity is g = 9.8 m/s^2.
Plugging in these values into the equation, we get:
s = (170 m/s)(5.8 s) + (1/2)(9.8 m/s^2)(5.8 s)^2
Calculating this equation will give us the displacement of the bullet after 5.8 seconds.
s = (170 m/s)(5.8 s) + (0.5)(9.8 m/s^2)(5.8 s)^2
s = 986 m + 0.5(9.8 m/s^2)(33.64 s^2)
s = 986 m + 0.5(322.932 m)
s = 986 m + 161.466 m
s = 1147.466 m
Therefore, the displacement of the bullet after 5.8 seconds, neglecting air resistance, is approximately 1147.466 meters.