A bullet is fired from a rifle that is held 2.5 m above the ground in a horizontal position. The initial speed of the bullet is 1450 m/s. Find:
(a) the time it takes for the bullet to strike the ground and
(b) the horizontal distance traveled by the bullet.
a. h = 0.5g*t^2 = 2.5 m.
4.9t^2 = 2.5 m.
t^2 = 0.51
Tf = 0.714 s.
b. Dx = Xo*Tf = 1450m/s * 0.714s=1026 m.
To find the time it takes for the bullet to strike the ground, we need to use the equation of motion for vertical motion:
\[ s = ut + \frac{1}{2}at^2 \]
where:
s = vertical displacement (drop in height)
u = initial velocity in the vertical direction
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken
In this case, the initial vertical velocity is zero (since the bullet is fired horizontally) and the displacement is -2.5 m (negative because the bullet is dropping).
\[ -2.5 = 0t + \frac{1}{2}(-9.8)t^2 \]
Simplifying, we get:
\[ 4.9t^2 = 2.5 \]
Dividing both sides by 4.9:
\[ t^2 = \frac{2.5}{4.9} \]
Taking the square root of both sides:
\[ t = \sqrt{\frac{2.5}{4.9}} \]
Calculate the value on the right-hand side to find the time it takes for the bullet to strike the ground.
To find the horizontal distance traveled by the bullet, we multiply the time taken by the initial horizontal velocity. Since the bullet is fired horizontally, the horizontal velocity remains constant throughout its motion.
The horizontal distance traveled is given by:
\[ d = u \cdot t \]
where:
d = horizontal distance traveled
u = initial horizontal velocity
t = time taken (as calculated in the previous step)
Substitute the known values to find the horizontal distance traveled by the bullet.