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].
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
1 answer