Question

A ship fires its guns with a speed of 400 m/s at an angle of 35 with the horizontal.
a. Find the range and maximum altitude.
b. If the ship hits a plane that was 12 km away, how high was the plane flying?

Answer 1

(a) Maximum height formula for projectile motion:

H,max = (vo,y)^2 / 2g
or
H,max = (vo)^2 * sin^2 (θ) / 2g
where
vo = initial velocity
θ = angle of release (wrt to horizontal)
g = acceleration due to gravity = 9.8 m/s^2

Substituting,
H,max = (400^2) * (sin (35))^2 / (2*9.8)
H,max = ?

(b) We are given a horizontal distance of 12 km = 12000 m, and we need to find the height of the bullet after it has traveled 12000 m. Formula to use:

x = (vo,x) * t
or
x = (vo cos(θ)) * t
where
x = horizontal distance
t = time

Solving for the time,
12000 = 400 cos(35) * t
t = 12000 / 327.661
t = 36.623 s

Then we use another formula to get the height, y:
y = (vo sin(θ))t - (1/2)gt^2

Substituting,
y = (400 sin(35))(36.623) - (0.5)(9.8)(36.623)^2
y = ?

Hope this helps~ `u`

Answer 2

I'm sorry, but I don't think part a's equation is right, unless I read the equation wrong. The answer key says, part a is 2.7 x 103 m = 278.1 and 1.5 x 104 m = 156. Also, for part b, if I read the equation right, is the answer 14,981.3562 m?