A bullet is fired from a rifle that is held 1.06 m above the ground in a horizontal position. The initial speed of the bullet is 1340 m/s. Find (a) the time it takes for the bullet to strike the ground and (b) the horizontal distance traveled by the bullet.
I am not really sure which formula to use.
To solve this problem, we can use kinematic equations, specifically the equations of motion in a vertical direction and a horizontal direction.
(a) To find the time it takes for the bullet to strike the ground, we need to calculate the time it takes for the bullet to reach the ground in the vertical direction.
Using the equation of motion in the vertical direction:
h = ut + (1/2)gt^2
where h is the initial height (1.06 m), u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Since the initial vertical velocity is zero (as the rifle is held horizontally), the equation simplifies to:
h = (1/2)gt^2
Rearranging the equation, we get:
t^2 = (2h) / g
t^2 = (2 * 1.06 m) / (-9.8 m/s^2)
Calculating t:
t^2 = -0.216 m / s^2
t = √(-0.216 m / s^2)
t ≈ 0.466 s (ignoring the negative value as time cannot be negative)
Therefore, it takes approximately 0.466 seconds for the bullet to strike the ground.
(b) To find the horizontal distance traveled by the bullet, we can use the equation of motion in the horizontal direction:
s = ut + (1/2)at^2
where s is the horizontal distance, u is the initial horizontal velocity (1340 m/s), a is the acceleration in the horizontal direction (which is zero since there is no horizontal acceleration), and t is the time.
Since there is no horizontal acceleration, the equation simplifies to:
s = ut
Plugging in the values:
s = (1340 m/s) * (0.466 s)
s ≈ 624.44 m
Therefore, the horizontal distance traveled by the bullet is approximately 624.44 meters.