A bullet is fired horizontally from a height of 78.4m and hits the ground 2400 m away. With what velocity does the bullet leave the ground?
To find the velocity at which the bullet leaves the ground, we can use the principles of projectile motion.
The horizontal distance traveled, in this case, is given as 2400 m, and the initial height is 78.4 m.
We need two equations to solve this problem:
1. Vertical motion equation:
Δy = V₀yt + (1/2)gt²
2. Horizontal motion equation:
Δx = V₀xt
In the vertical motion equation, Δy represents the vertical displacement, V₀y is the vertical component of the initial velocity (which is zero because the bullet is fired horizontally), g is the acceleration due to gravity (approximately 9.8 m/s²), t is the time of flight.
In the horizontal motion equation, Δx represents the horizontal displacement, V₀x is the horizontal component of the initial velocity, and t is the time of flight.
Since the bullet is fired horizontally, the vertical component of the initial velocity is zero (V₀y = 0). Therefore, we can ignore the vertical motion equation for now.
Using the horizontal motion equation:
Δx = V₀xt
2400 m = V₀x * t
To find t, we need to calculate the time of flight, which can be done using the equation:
t = Δy / V₀y
t = 78.4 m / 0 m/s (since V₀y = 0 for the horizontal projectile)
Now substitute the value of t in the equation Δx = V₀x * t:
2400 m = V₀x * (78.4 m / 0 m/s)
Since t = 0 m/s, the equation simplifies to:
2400 m = V₀x * 0
This implies that the horizontal component of the initial velocity (V₀x) is indeterminate. The bullet can leave the ground with any horizontal velocity since it doesn't depend on the time of flight.
how long does it take to fall 78.4m?
h=1/2 g t^2
in that time,
velocity must be = 2400/time