an archer at the edge of a cliff fires an arrow from a height of 14.6 m above the ground at the bottom of the cliff. the arrow just clears a rock on a pillar that is 21.7 m above the gound which is 45.2 m horizontally away from the archer.determine
a,the initial speed of the arrow
b,the angle above the horizontal that the arrow is fired at
c,how far past the rock on the pillar that the arrow lands
To solve this problem, we can use the principles of projectile motion. We'll break down the motion into vertical and horizontal components.
a) To find the initial speed of the arrow, we can use the information given and the equations of motion. The vertical component of the motion is influenced by gravity, so we can use the following equation:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy = change in vertical displacement (21.7 m - 14.6 m = 7.1 m)
v₀y = initial vertical component of velocity
t = time of flight
a = acceleration due to gravity (-9.8 m/s²)
Since we want to solve for v₀y, we can consider the following equation:
v₀y = (Δy - (1/2) * a * t²) / t
Now, let's find the time of flight. The projectile is fired horizontally, so the time taken for the horizontal and vertical components will be the same. We can use the following equation:
Δx = v₀x * t
Where:
Δx = horizontal displacement (45.2 m)
v₀x = initial horizontal component of velocity
t = time of flight
Since the horizontal component of the velocity remains constant throughout the motion, we can use the equation:
v₀x = Δx / t
Given that Δx = 45.2 m, we need to find the value of t to find v₀x.
To find t, we can use the vertical motion equation:
Δy = v₀y * t + (1/2) * a * t²
Given that Δy = 7.1 m and a = -9.8 m/s², we need to rearrange the equation to find t.
First, multiply both sides by 2 to eliminate the fraction:
2 * Δy = 2 * (v₀y * t) + (1/2) * a * t²
Simplify further:
2 * Δy = 2v₀y * t + (1/2) * a * t²
Rearrange the equation:
(1/2) * a * t² + 2v₀y * t - 2 * Δy = 0
Now, we have a quadratic equation in terms of t. We can use the quadratic formula to solve for t.
t = (-b ± √(b² - 4ac)) / 2a
Where:
a = (1/2) * a
b = 2v₀y
c = -2 * Δy
Using the values, plug them into the quadratic formula and solve for t.
Once we have the value of t, we can substitute it into the equation for v₀x to find the initial horizontal component of the velocity.
b) To find the angle above the horizontal at which the arrow is fired, we'll use the value of v₀y found in part a.
The angle can be determined using trigonometric functions:
θ = tan⁻¹(v₀y / v₀x)
Substitute the values of v₀y and v₀x to calculate the angle.
c) Finally, to find how far past the rock on the pillar the arrow lands, we need to calculate the horizontal distance covered by the arrow. This can be done using the formula:
Δx = v₀x * t
Substitute the values of v₀x and t to find the horizontal distance.