a bullet is fired from a gun that is 2.0 meters above flat ground. the bullet emerges from the gun with a speed of 250 m/s.
a) how long does the bullet stay in the air before it hits the ground?
b) At what horizonatal distance from the firing point does the bullet strike the ground?
i know i need to do horizontal and vertical components, but after that im lost.
90
To solve this problem, we need to split the motion of the bullet into its vertical and horizontal components.
a) Vertical Motion:
The vertical component of motion is influenced by gravity. We will use the equation:
h = ut + (1/2)gt^2
where
h = vertical displacement (in this case, the bullet's initial position above the ground)
u = initial vertical velocity (which is 0 because the bullet is not moving vertically at first)
g = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time
In this case, we need to solve for t when h = 2.0 meters. Rearranging the equation, we get:
(1/2)gt^2 = h - ut
(1/2)gt^2 = 2.0 - 0
(1/2)gt^2 = 2.0
gt^2 = 4.0
t^2 = 4.0 / g
t = √(4.0 / g)
Now we can substitute the value of g into the equation to find the time it takes for the bullet to hit the ground.
b) Horizontal Motion:
The horizontal motion of the bullet is unaffected by gravity unless there is air resistance (which we will neglect in this case). The horizontal component of motion is given by the equation:
d = vt
where
d = horizontal displacement
v = horizontal velocity (which is constant)
t = time
In this case, we need to solve for d (the horizontal distance the bullet travels) when v = 250 m/s (given). We will use the time calculated in part a), which is the time of flight of the bullet.
Now we can calculate the horizontal distance the bullet travels using the equation d = vt.
By applying these equations, we can find the answers to both parts of the question.