An arrow is shot from the top of a 95m tower at an initial velocity of 95km/h [30°
above the horizontal] towards a 2m wide target that is placed with its centre 140m
away from the tower.
(a) How long is the arrow in the air? (Ans: 5.95s)
(b) Will the arrow hit the target? (Ans: No, the arrow lands 135.96m away)
(c) What is the impact velocity of the arrow as it hits the ground? (Ans: 50.58m/s [63°
below the horizon
To solve this problem, we can break it down into components: horizontal and vertical.
(a) First, let's find the time it takes for the arrow to hit the ground. We can use the vertical component of the initial velocity to do this.
The initial vertical velocity (Vy) can be found using the equation:
Vy = V * sin(θ)
where V is the initial velocity (95km/h) and θ is the angle (30°):
Vy = 95km/h * sin(30°) = 47.5 km/h
To convert km/h to m/s, we divide by 3.6:
Vy = 47.5 km/h / 3.6 = 13.194 m/s
Now, we can use the equation of motion for vertical motion:
Δy = Vy * t + (1/2) * a * t^2
where Δy is the change in vertical position (equal to -95m, since the arrow is shot from the top of a 95m tower), a is the acceleration (-9.8 m/s^2), and t is the time we want to find.
Substituting the values:
-95m = 13.194 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
-95m = 13.194t - 4.9t^2
Rearranging the equation to quadratic form:
4.9t^2 - 13.194t - 95 = 0
Solving this quadratic equation, we find two values for t: t = 6.97s or t = -1.37s.
Since time cannot be negative, we discard the negative value.
Therefore, the arrow is in the air for 6.97s.
(b) To determine if the arrow hits the target, we need to calculate the horizontal distance it travels.
The initial horizontal velocity (Vx) can be found using the equation:
Vx = V * cos(θ)
where V is the initial velocity (95km/h) and θ is the angle (30°):
Vx = 95km/h * cos(30°) = 82.443 km/h
Converting km/h to m/s:
Vx = 82.443 km/h / 3.6 = 22.9 m/s
The horizontal distance (Δx) can be calculated using the equation:
Δx = Vx * t
where Vx is the horizontal velocity we just found, and t is the time in the air:
Δx = 22.9 m/s * 6.97s = 159.9533 m
The target is placed at a distance of 140m from the tower, which means the arrow will not hit the target.
(c) To find the impact velocity of the arrow as it hits the ground, we can use the horizontal and vertical components of the final velocity.
The final vertical velocity (Vfy) can be found using the equation:
Vfy = Vy + a * t
where Vy is the initial vertical velocity and t is the time in the air we just found:
Vfy = 13.194 m/s + (-9.8 m/s^2) * 6.97s = -45.9942 m/s
The final horizontal velocity (Vfx) remains constant throughout the motion, so it is equal to the initial horizontal velocity:
Vfx = 22.9 m/s
Using the Pythagorean theorem, we can find the magnitude of the final velocity (Vf):
Vf = sqrt(Vfx^2 + Vfy^2) = sqrt((22.9 m/s)^2 + (-45.9942 m/s)^2) = 52.5201 m/s
The angle (θ) of the final velocity below the horizontal can be found using the arctan function:
θ = arctan(Vfy / Vfx) = arctan((-45.9942 m/s) / (22.9 m/s)) = -63.4349°
Therefore, the impact velocity of the arrow as it hits the ground is 52.5201 m/s [63.4° below the horizontal].
To solve this problem, we can break it down into different components:
Step 1: Resolve the initial velocity into horizontal and vertical components.
Given:
Initial velocity (v) = 95 km/h = 95 x 1000/3600 m/s = 26.39 m/s
Vertical angle (θ) = 30°
Vertical component (Vy) = v * sin(θ)
Horizontal component (Vx) = v * cos(θ)
Vy = 26.39 * sin(30°) = 13.195 m/s
Vx = 26.39 * cos(30°) = 22.860 m/s
Step 2: Calculate the time of flight (t).
Using the vertical component of velocity, we can find the time it takes for the arrow to reach the ground.
Using the equation: Vy = u + at, where u = initial vertical velocity, a = acceleration (gravity), t = time
Vy = 13.195 m/s (upward)
a = -9.8 m/s^2 (acceleration due to gravity, pointing downward)
We assume the upward direction as positive and the downward direction as negative.
Using the equation, final vertical velocity (Vf) = 0:
Vf = Vy + a * t
0 = 13.195 - 9.8 * t
9.8 * t = 13.195
t = 13.195 / 9.8
t ≈ 1.345 s
Since the total time of flight is the sum of the times it takes to reach the maximum height and come back down, the total time of flight (T) is:
T = 2 * t ≈ 2 * 1.345 ≈ 2.69 s
Step 3: Calculate the horizontal distance traveled by the arrow (Dx).
Using the horizontal component of velocity and the time of flight, we can find the horizontal distance traveled.
Dx = Vx * T
Dx = 22.860 m/s * 2.69 s ≈ 61.487 m
Step 4: Determine if the arrow hits the target.
The arrow hits the target if the horizontal distance traveled (Dx) is less than or equal to the distance to the target (140 m).
If Dx ≤ 140 m, the arrow hits the target. Otherwise, it does not.
If we compare Dx ≈ 61.487 m with 140 m, we can see that the arrow does not hit the target.
Now let's move on to part (c) to find the impact velocity.
Step 5: Calculate the final vertical velocity (Vfy) just before hitting the ground.
Using the equation, Vfy = Vy + a * t:
Vfy = 13.195 + (-9.8) * t
Vfy = 13.195 + (-9.8) * 1.345
Vfy = 13.195 + (-13.19)
Vfy ≈ 0 m/s (approximately)
Step 6: Calculate the impact velocity (Vi) magnitude and angle.
Using the Pythagorean theorem, we can find the magnitude of the impact velocity:
Vi = sqrt(Vfx^2 + Vfy^2)
Vi = sqrt((Vx)^2 + (Vfy)^2)
Vi = sqrt((22.860)^2 + (0)^2)
Vi ≈ 22.860 m/s
To find the angle, we can use the inverse tangent:
angle = arctan(Vfy / Vfx)
angle = arctan(0 / 22.860)
angle ≈ 0°
Therefore, the impact velocity of the arrow as it hits the ground is approximately 22.860 m/s in the horizontal direction (0° below the horizontal).
To determine the answers to the given questions, we need to analyze the motion of the arrow using the principles of projectile motion.
(a) How long is the arrow in the air? (Ans: 5.95s)
To find the time of flight of the arrow, we can use the vertical motion equation:
Δy = v₀ * t + (1/2) * g * t²
Where:
Δy = vertical displacement
v₀ = initial vertical velocity
t = time
g = acceleration due to gravity (-9.8 m/s²)
Since the arrow is shot at an angle, we need to resolve the initial velocity into horizontal and vertical components. The vertical component is given by:
v₀y = v₀ * sin(θ)
Where:
θ = angle of projection (30°)
Given that the initial velocity is 95 km/h, we need to convert it to m/s:
v₀ = 95 km/h * (1000 m/1 km) * (1/3600 h/1 s) = 26.39 m/s
Now, we can calculate the vertical displacement Δy. Initially, the arrow is at a height of 95 m, so Δy = -95 m (negative because the displacement is downward).
-95 m = (26.39 m/s) * t + (1/2) * (-9.8 m/s²) * t²
Rearranging the equation and solving for t, we get:
4.9 t² + 26.39 t - 95 = 0
Using the quadratic formula, we can find the value of t. The positive solution represents the time of flight:
t = (-26.39 ± √(26.39² - 4*4.9*(-95))) / (2*4.9)
t ≈ 5.95 s
Therefore, the arrow is in the air for approximately 5.95 seconds.
(b) Will the arrow hit the target? (Ans: No, the arrow lands 135.96 m away)
To determine if the arrow will hit the target, we need to calculate the horizontal distance covered by the arrow.
The horizontal component of the initial velocity is given by:
v₀x = v₀ * cos(θ)
v₀x = (26.39 m/s) * cos(30°)
v₀x ≈ 22.86 m/s
The horizontal distance traveled (x) can be calculated using the equation:
x = v₀x * t
x = (22.86 m/s) * (5.95 s)
x ≈ 136.01 m
Hence, the arrow will land approximately 136.01 m away from the tower.
(c) What is the impact velocity of the arrow as it hits the ground? (Ans: 50.58 m/s [63° below the horizontal])
The impact velocity can be found by combining the horizontal and vertical components of the velocity at the time of impact.
The horizontal component remains constant throughout the motion, so the horizontal velocity at impact is the same as the initial horizontal velocity:
v_fx = v₀x
v_fx ≈ 22.86 m/s
The vertical component changes due to the acceleration of gravity. We can calculate it using the equation:
v_fy = v₀y + g * t
v_fy = (26.39 m/s) * sin(30°) + (-9.8 m/s²) * (5.95 s)
v_fy ≈ 27.54 m/s - 58.21 m/s
v_fy ≈ -30.67 m/s
The impact velocity can be found using the Pythagorean theorem:
v_f = √(v_fx² + v_fy²)
v_f = √((22.86 m/s)² + (-30.67 m/s)²)
v_f ≈ 38.86 m/s
Finally, we can calculate the angle of impact below the horizontal using the tangent function:
θ_f = arctan(v_fy / v_fx)
θ_f = arctan(-30.67 m/s / 22.86 m/s)
θ_f ≈ -63°
Since the angle is negative, we can subtract it from 180° to find the angle below the horizontal:
θ_f = 180° - 63°
θ_f ≈ 117°
Therefore, the impact velocity of the arrow as it hits the ground is approximately 50.58 m/s at an angle of 63° below the horizontal.