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.
Find the bullet's range?
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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
vx = v0= 280 m/s.
For the y-component we have the kinematic equation
vy=-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 = √(vx²+vy²) = √(280²+(-54.2)²) = 285 m/s.
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 :)