Two boys simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with a speed of 100km/s at an angle of projection of 30. The second boy is ahead of the first boy by a distance of 50m and releases his shot with a speed of 80m/s. how must he aim his gun so that both the shots hit the bird simultaneously? What is the distance of the foot of the tower from the two boys and the height of the tower? With what velocities and when do the two shots hit the bird?
To determine how the second boy must aim his gun so that both shots hit the bird simultaneously, we need to calculate the time it takes for each shot to reach the bird.
First, let's calculate the time it takes for the first boy's shot to reach the bird. We have the initial speed (v1) of 100 km/s and the angle of projection (θ1) of 30 degrees. We'll use the following equations of projectile motion:
Vertical motion equation:
y = v1 * t * sin(θ1) - (1/2) * g * t^2
Horizontal motion equation:
x = v1 * t * cos(θ1)
Given that y = 0 (the bird is at the same height as the tower) and x = 50 m (the horizontal distance between the first boy and the second boy), we can solve for t:
0 = v1 * t * sin(θ1) - (1/2) * g * t^2 -- (Equation 1)
50 = v1 * t * cos(θ1) -- (Equation 2)
We know that the acceleration due to gravity (g) is approximately 9.8 m/s^2.
From Equation 2, we can isolate t:
t = 50 / (v1 * cos(θ1)) -- (Equation 3)
Now, let's calculate the time it takes for the second boy's shot to reach the bird. We have the initial speed (v2) of 80 m/s, and we need to find the angle of projection (θ2). Using the same equations of projectile motion, we can set up the following equations:
0 = v2 * t * sin(θ2) - (1/2) * g * t^2 -- (Equation 4)
50 + x = v2 * t * cos(θ2) -- (Equation 5)
Substituting the value of t from Equation 3 into Equation 5, we can solve for θ2:
50 + x = v2 * (50 / (v1 * cos(θ1))) * cos(θ2)
θ2 = arccos[(50 + x) * (v1 * cos(θ1)) / (50 * v2)] -- (Equation 6)
We can use this equation to find the angle θ2.
Now that we have the values of θ2, we can calculate the distance of the foot of the tower from the two boys and the height of the tower.
Distance of the foot of the tower (d):
Using the horizontal motion equation (x = v1 * t * cos(θ1)), we can calculate the distance d by substituting the values of v1, t, and θ1.
Height of the tower (h):
Using the vertical motion equation (y = v1 * t * sin(θ1) - (1/2) * g * t^2), we can calculate the height h by substituting the values of v1, t, θ1, and g.
Finally, to find the velocities and timing of both shots hitting the bird, we can use the equations of projectile motion with the respective θ1 and θ2 values, as well as the calculated time t.
Please note that I've made an assumption that the bird is at the same height as the tower. If this assumption is incorrect, please provide the height of the tower for a more accurate solution.
To solve this problem, we can break it down into two main steps:
Step 1: Find the time it takes for the first boy's shot to reach the bird.
Step 2: Use the time calculated in Step 1 to find the angle and other parameters for the second boy's shot.
Let's start with Step 1:
1. Find the time for the first boy's shot to reach the bird:
- We know that the initial speed of the first boy's shot is 100 km/s, and the angle of projection is 30 degrees.
- We can break down the initial speed into its horizontal and vertical components:
- Horizontal component: Vx1 = 100 km/s * cos(30)
- Vertical component: Vy1 = 100 km/s * sin(30)
- The bird is initially at rest, so its vertical position is 0.
- We can use the equation for vertical displacement to find the time (t1) it takes for the first boy's shot to reach the bird's height:
- Vertical displacement equation: y = Vy1 * t1 - (1/2) * g * t1^2
- The vertical displacement (y) is the height of the tower.
- The acceleration due to gravity (g) is approximately 9.8 m/s^2.
- Solve the equation for t1.
Now that we have the time t1, we can move on to Step 2:
2. Find the angle and other parameters for the second boy's shot:
- The second boy is ahead of the first boy by a distance of 50 m.
- We need to calculate the horizontal distance traveled by the bird in time t1.
- Horizontal distance equation: x = Vx1 * t1 + (1/2) * ax * t1^2
- The bird is initially at rest, so its horizontal position is 0.
- The acceleration in the x-direction (ax) is 0 since there are no external forces acting horizontally.
- Solve the equation for x.
- Now that we know the horizontal distance from the second boy to the bird, we can calculate the angle (θ2) at which the second boy must aim his gun.
- The horizontal distance forms the base of a right triangle.
- The height of the triangle is the 50 m distance between the two boys.
- The angle θ2 is the arctan(height / base).
- Finally, we can calculate the vertical and horizontal components of the second boy's initial velocity (Vx2 and Vy2) using the given speed of 80 m/s and the calculated angle θ2.
With the times of flight and the parameters calculated, you can determine the distance of the foot of the tower from the two boys and the height of the tower. The velocities and times when both shots hit the bird will also be available based on the calculations above.
Suppose that the bullet hits the aim at the upward motion, then
h= v₀₁•sinα•t - gt²/2 =
v₀₂•sinβ•t - gt²/2,
v₀₁•sinα = v₀₂•sinβ ,
sinβ= v₀₁•sinα/ v₀₂=
=100•sin30⁰/80 =0.625,
β=sin⁻¹0.625 = 38.7⁰.
x₁-x₂=Δx,
v₀₁•cosα•t = v₀₂•cosβ•t = Δx,
t= Δx/{ v₀₁•cosα - v₀₂•cosβ}=
=50/{100•cos30⁰ -80•cos38.7⁰}=
=2.07 s.
h=v₀₁•sinα•t - gt²/2=
=100•0.5•2.07 = 9.8•2.07²/2 =82.5 m.
x₁=v₀₁•cosα•t =100•cos30⁰•2.07 =179.3 m
x₂=x₁-Δx = 179.3-50 =129.3 m.
v₁(y)=v₀₁•sinα - g•t =
=100•0.5 – 9.8•2.07 =29.7 m/s.
v₁(x)= v₀₁•cosα =100•0.866 =86.6 m/s,
v₁= sqrt{v₁(x)²+v₁(y)²} =
=sqrt{29.7²+86.6²} = 91.6 m/s.
v₂(y)=v₀₂•sinβ - g•t =
=80•0.625 – 9.8•2.07 =29.7 m/s.
v₂ (x)= v₀₂•cosβ =80•0.78 =62.4 m/s,
v₂= sqrt{v₂ (x)²+v₂ (y)²} =
=sqrt{29.7²+62.4²} = 69.1 m/s.