Determine the launch speed of a horizontally-launched projectile which lands 27.4 meters from the base of a 17.9-meter high cliff.
To determine the launch speed of a horizontally-launched projectile, we can use the equations of motion.
First, let's analyze the vertical motion of the projectile. We can use the equation:
y = v₀t + (1/2)gt²
where:
y = vertical displacement (negative because it's downward)
v₀ = initial vertical velocity (which is 0 m/s for a horizontally-launched projectile)
t = time of flight (unknown)
g = acceleration due to gravity (approximately 9.8 m/s²)
Since the projectile lands 17.9 meters above its launch point, we can rewrite the equation as:
-17.9 = 0(t) + (1/2)(9.8)(t²)
Simplifying the equation, we get:
4.9t² = 17.9
Dividing both sides by 4.9, we have:
t² = 3.653
Taking the square root of both sides:
t ≈ 1.91 seconds
Note that this time represents the total time of flight, considering both the upward and downward motion of the projectile.
Now, let's analyze the horizontal motion. The horizontal displacement is 27.4 meters, and the time of flight is 1.91 seconds. We can use the equation:
x = v₀x * t
where:
x = horizontal displacement (27.4 meters)
v₀x = initial horizontal velocity (unknown)
t = time of flight (1.91 seconds)
Rearranging the equation, we get:
v₀x = x / t
Substituting the values:
v₀x = 27.4 / 1.91
v₀x ≈ 14.35 m/s
Therefore, the launch speed of the horizontally-launched projectile is approximately 14.35 m/s.
To determine the launch speed of a horizontally-launched projectile, we can use the kinematic equation for horizontal motion:
d = v * t
where:
- d is the horizontal distance traveled by the projectile
- v is the launch speed of the projectile
- t is the time of flight
In this case, we need to find the launch speed (v).
We have two key pieces of information:
1. The projectile lands at a horizontal distance of 27.4 meters from the base of the cliff.
2. The cliff is 17.9 meters high.
To find the time of flight, we can use the equation for vertical motion under constant acceleration:
h = (1/2) * g * t^2
where:
- h is the height of the cliff (17.9 meters)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time of flight
Rearranging the equation, we get:
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 17.9) / 9.8)
t ≈ sqrt(3.653)
Therefore, the time of flight is approximately t ≈ 1.91 seconds.
Now that we have the time of flight, we can find the launch speed (v) using the horizontal distance (d) traveled by the projectile:
d = v * t
Rearranging the equation, we get:
v = d / t
Substituting the given values:
v = 27.4 meters / 1.91 seconds
Therefore, the launch speed of the horizontally-launched projectile is approximately v ≈ 14.36 m/s.