A cannon is shooting emergency packets to people stranded on the roof of a flooded building of height H = 104 meters relative to the cannon, the corner of which is located a distance D = 53 meters from the cannon. It is desired that the incoming packets are flying tangent to the roof as shown so that they land gently with as little impact as possible and slide along to a stop.
Find the initial speed v0 and at what angle è (in radians) the cannon should be aimed to achieve the above scenario.
vertical problem (time and speed same as 104 meter fall):
v = Vi - 9.81 t
0 = Vi - 9.81 t
t = Vi/9.81
104 = (1/2) g t^2
104 = (1/2) (9.81)(Vi^2/9.81^2)
Vi^ = 2 * 9.81 * 104
Vi = 45.2 m/s
horizontal problem (constant speed)
t = Vi/9.81 = 4.61 seconds
u = 53/4.61 = 11.5 m/s
tan (angle) = 45.2/11.5
so angle up from horizontal = 75.7 degrees
speed = sqrt (45.2^2 + 11.5^2)
= 46.6 m/s
To find the initial speed v₀ and the angle θ at which the cannon should be aimed, we can use the principles of projectile motion.
Let's break down the problem step by step:
Step 1: Decompose the initial velocity:
Since we need the packets to land gently on the roof, we can assume that the vertical component of the velocity (v_y) at impact is zero. Therefore, the initial velocity (v₀) can be decomposed into horizontal (v₀x) and vertical (v₀y) components:
v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)
Step 2: Calculate the time of flight:
The time of flight (T) can be calculated using the equation:
T = (2 * v₀y) / g
Where g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 3: Calculate the range:
The range (R) is the horizontal distance traveled by the packet. It is given by:
R = v₀x * T
Step 4: Find the point of impact:
The point of impact should be at a horizontal distance D from the cannon. Therefore, by setting the range (R) equal to D, we can solve for v₀x:
v₀x * T = D
v₀x * [(2 * v₀y) / g] = D
v₀ * cos(θ) * [(2 * v₀ * sin(θ)) / g] = D
Simplifying, we find:
v₀² * sin(2θ) = (g * D) / 2
Step 5: Find the initial speed:
Now that we have an equation relating v₀ and θ, we can plug in the values of D, g, and solve for v₀. Since v₀ is positive (speed), we can ignore any negative solutions:
v₀² = [(g * D) / 2] / sin(2θ)
v₀ = sqrt([(g * D) / 2] / sin(2θ))
Step 6: Find the angle θ:
Finally, we can calculate the angle θ by rearranging the equation solved for v₀:
θ = 0.5 * arcsin([(g * D) / 2] / v₀²)
Substitute the given values of H = 104 meters and D = 53 meters into the above equation to find the angle θ. Then substitute the calculated value of θ back into the equation for v₀ to find the initial speed v₀.
Note: Make sure to convert the angle θ to radians when using trigonometric functions.