a bullet is fired from the ground at the angle of 30 degree above the horizontal,What initial speed must the bullet have in order to hit a point 250 ft high on a tower located 400 ft away (ignoring air resistance),what the solution?
Write vertical and horizontal equations of motion. Require that X = 400 and Y = 250 at same time, t. You have two equations for the two unknowns, t and Vo. They are:
X = Vo*cos30 * t = 400
Y = Vo*sin30 * t - (g/2) t^2 = 250
I recommend using the X equation to substitute for t in the second (Y) equation.
g = 32.2 ft/s^2.
To find the initial speed of the bullet, we can use the equations of motion for projectile motion. Here's how we can approach this problem:
Step 1: Understand the problem
We are given the angle of elevation (30 degrees), the height of the tower (250 ft), and the horizontal distance to the tower (400 ft). We need to find the initial speed of the bullet.
Step 2: Analyze the motion
Since the bullet is fired at an angle above the horizontal, it will have both horizontal and vertical components of velocity. The vertical component will cause the bullet to rise to a certain height and then fall back down, while the horizontal component will make the bullet travel a certain distance. The time of flight will be the same for both vertical and horizontal motion.
Step 3: Break it down into components
Given that the angle of elevation is 30 degrees, we can calculate the initial vertical and horizontal velocities:
Vertical component: Vy = V * sin(θ)
Horizontal component: Vx = V * cos(θ)
Where V is the initial velocity of the bullet and θ is the angle of elevation (30 degrees).
Step 4: Calculate the time of flight
To calculate the total time of flight, we need to use the vertical component of motion since the height of the tower is in the vertical direction. We can use the following equation:
h = (Vy * t) - (0.5 * g * t^2)
Where h is the height of the tower (250 ft), t is the time of flight, and g is the acceleration due to gravity (32.2 ft/s^2).
Since the bullet will hit the tower when it reaches its maximum height, the initial and final vertical velocities will be equal, so we have:
0 = (Vy * t) - (0.5 * g * t^2)
Step 5: Solve for t
Rearranging the equation, we get:
0.5 * g * t^2 = (Vy * t)
0.5 * 32.2 * t^2 = (V * sin(θ) * t)
16.1 * t = V * sin(θ)
Step 6: Calculate the horizontal distance
To find the horizontal distance traveled by the bullet, we can use the horizontal component of velocity (Vx) and the total time of flight (t):
horizontal distance = Vx * t
horizontal distance = (V * cos(θ)) * t
We know that the horizontal distance is 400 ft, so we can write the equation as:
400 = V * cos(θ) * t
Step 7: Solve the system of equations
Now, we have two equations with two unknowns: V and t. We can solve the system of equations simultaneously. Substitute the value of t from the horizontal distance equation into the equation derived from the vertical motion:
400 = V * cos(θ) * (16.1 * t / V * sin(θ))
Simplifying further:
400 = 16.1 * cos(θ) / sin(θ)
Step 8: Solve for V
Rearranging the equation:
V = 400 * sin(θ) / (16.1 * cos(θ))
Substituting the value of θ (30 degrees) and evaluating the expression:
V = 400 * sin(30) / (16.1 * cos(30))
V = 400 * 0.5 / (16.1 * 0.866)
V = 100 / 13.92
V ≈ 7.18 ft/s
So, the initial speed of the bullet must be approximately 7.18 ft/s in order to hit a point 250 ft high on a tower located 400 ft away.