Find the bullet's range?

Question

A bullet was fired horizontally from the top of a 150 m tall tower at 280 m/s. Find the bullet's range and time of flight. Determine also its speed on impact with the ground below.

Answer 1

We have the kinematic equations

 

Δx = v₀t,  Δy = -½gt²

 

With Δy = -150 m, solving the second equation for t gives us

 

t = √(2(150)/9.8) s = 5.53 s.

 

This is the flight time. Plug it into the first equation to get the range:

 

Δx =280 (5.53) m = 1550 m.

 

To find the speed on impact we need the two components of the final velocity. The velocity's x-component doesn't change, so

 

v_x = v₀= 280 m/s.

 

For the y-component we have the kinematic equation

 

v_y=-gt = -9.8 (5.53) m/s = -54.2 m/s

 

The final speed is the magnitude of the final velocity. Use the Pythagorean theorem to find

 

v = √(v_x²+v_y²) = √(280²+(-54.2)²) = 285 m/s.

Answer 2

 

Sun -- Vdn from KE=PE ... 2gh= V*V ... Vdn = sqrt 3000 ~ sqrt(60x50) ~ 55m/s ==> means 5.5s airtime ... range is 5*280 with 10% tip ~2800/2 ~ 1400+140 = 1540m range ... the Vdn part is a mere 1/5 of Vx, so Vf is essentially 280+m/s ... Regards :)