A cannon sends a projectile towards a target
a distance 1250 m away. The initial velocity
makes an angle 40◦ with the horizontal. The
target is hit.
The acceleration of gravity is 9.8 m/s2 .
What is the magnitude of the initial veloc-
ity?
Answer in units of m/s.
is that sin square?
d = V^2(sin^2(µ)/2g
d = 1250m
µ = 40º
g = 9.8m^s^2
Solve for V
To find the magnitude of the initial velocity, we can use the horizontal and vertical components of the velocity.
Given:
Distance to the target (d) = 1250 m
Launch angle (θ) = 40 degrees
Acceleration due to gravity (g) = 9.8 m/s^2
We can first find the time it takes for the projectile to reach the target.
Step 1: Horizontal component of velocity
The horizontal component of velocity (Vx) remains constant throughout the motion.
Vx = V * cos(θ)
Step 2: Vertical component of velocity
The vertical component of velocity (Vy) changes due to the acceleration of gravity.
Vy = V * sin(θ) - g * t
Since the target is hit, the vertical displacement (Δy) is 0.
Δy = Vy * t - (1/2) * g * t^2 = 0
Step 3: Solving for time (t)
Using the formula above, we can solve for t in terms of V.
(1/2) * g * t^2 = Vy * t
(1/2) * g * t = Vy
t = 2 * Vy / g
Step 4: Solving for the magnitude of the initial velocity (V)
Since V = sqrt(Vx^2 + Vy^2), we can substitute the values of Vx and Vy.
V = sqrt((V * cos(θ))^2 + (V * sin(θ) - g * t)^2)
Simplifying the equation, we get:
V = sqrt(V^2 * cos^2(θ) + (V * sin(θ) - g * t)^2)
Now, substitute the expression for t from step 3:
V = sqrt(V^2 * cos^2(θ) + (V * sin(θ) - g * (2 * Vy / g))^2)
Simplifying further:
V = √(V^2 * cos^2(θ) + (V * sin(θ) - 2 * Vy)^2)
Step 5: Solve for V using a numerical method or approximation technique.
Unfortunately, it is not possible to solve this equation analytically. We need to use numerical methods or approximation techniques to find the value of V.
To find the magnitude of the initial velocity, we can use the equations of projectile motion. We know that the horizontal motion and the vertical motion of the projectile are independent of each other. Let's break the initial velocity into its horizontal and vertical components.
Given:
Distance to the target (horizontal distance) = 1250 m
Angle with the horizontal = 40°
Acceleration due to gravity = 9.8 m/s²
Step 1: Find the horizontal component of the initial velocity.
The horizontal component of velocity (Vx) will remain constant throughout the motion. It can be found using the equation: Vx = V * cos(theta), where V is the initial velocity and theta is the launch angle.
Vx = V * cos(40°)
Step 2: Find the vertical component of the initial velocity.
The vertical component of velocity (Vy) will change due to the acceleration of gravity. It can be found using the equation: Vy = V * sin(theta) - g * t, where g is the acceleration due to gravity and t is the time of flight (which we assume to be the same time taken to reach the target).
Vy = V * sin(40°) - (9.8 * t)
Step 3: Relate the horizontal and vertical components of velocity.
Since the projectile hits the target, the vertical displacement (change in height) must be zero. Therefore, the time of flight can be found by solving the equation: 0 = Vy * t - (1/2) * g * t².
0 = (V * sin(40°) - (9.8 * t)) * t - (1/2) * 9.8 * t²
Step 4: Solve for t.
Rearrange the equation and solve for t using algebraic methods or numerical approximation techniques like the quadratic formula or numerical solvers.
Step 5: Substitute t back into the equation for Vy to find its value.
Vy = V * sin(40°) - (9.8 * t)
Step 6: Substitute the value of t and Vy into the equation for Vx to find the value of V.
Vx = V * cos(40°)
Once you have the horizontal and vertical components of the initial velocity (Vx and Vy), you can find the magnitude of the initial velocity (V) using the Pythagorean theorem:
V = √(Vx² + Vy²)
Plug in the values of Vx and Vy calculated in the previous steps to find the magnitude of the initial velocity in m/s.