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?

Tutors, please sign in to answer this question.

# 2 Answers

We have the kinematic equations

Δx = v

_{0}t, Δy = -½gt²With 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 }= v_{0}= 280 m/s.For the y-component we have the kinematic equation

v

_{y}=-gt = -9.8 (5.53) m/s = -54.2 m/sThe final speed is the magnitude of the final velocity. Use the Pythagorean theorem to find

v = √(v

_{x}²+v_{y}²) = √(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 :)