a projectile is fired at an angle of 27 above the horizontal from the top of a cliff 500 ft high. The initial speed of projectile is 1670 ft/s. How far will it move horizontally before hitting the level ground at the base of the cliff?
g = about 32 feet/second^2 as I remember from my childhood
so (1/2) g = 16
vertical problem
Vi = 1670 sin 27
Hi = 500
h = Hi + Vi t - 16 t^2
0 = 500 + (1670 sin 27) t - 16 t^2
solve quadratic for t, use the positive answer, that is time in the air
horizontal problem
u = 1670 cos 27 forever
range = u t
To solve this problem, we can use the equations of motion for projectile motion.
1. First, we need to break down the initial velocity into its horizontal and vertical components.
The vertical component (Vy) can be calculated using the initial speed (Vi) and the angle (θ) as follows:
Vy = Vi * sin(θ)
Substituting the given values:
Vy = 1670 ft/s * sin(27°)
2. Next, we can determine the time of flight (t) when the projectile hits the ground. The time of flight for a projectile launched from an elevated position can be found using the equation:
t = (2 * Vy) / g
where g is the acceleration due to gravity, approximately 32.2 ft/s^2.
Substituting the given values:
t = (2 * 1670 ft/s * sin(27°)) / 32.2 ft/s^2
3. Once we have the time of flight, we can find the horizontal distance traveled (X) using the equation:
X = Vx * t
where Vx is the horizontal component of the initial velocity, which can be calculated as:
Vx = Vi * cos(θ)
Substituting the given values:
Vx = 1670 ft/s * cos(27°)
Finally, substituting these values into the equation for X:
X = (1670 ft/s * cos(27°)) * [(2 * 1670 ft/s * sin(27°)) / 32.2 ft/s^2]
Evaluating this expression will give you the horizontal distance traveled by the projectile before hitting the ground.