After landing safely on the target the cat tries another projectile apparatus. This time the cat is shot out of a cannon over a 30 m high wall. The cat is launched at an angle of 55º0 and can be assumed to be at ground level during launch. With what speed (in km/h) does it have to be launched to make it approximately 5 m over the wall if the wall is 250 m from the cannon?
To solve this problem, we can use the principles of projectile motion. We need to find the launch speed that will allow the cat to clear a 30 m high wall and land approximately 250 m away.
Let's break down the given information:
- Initial vertical displacement: The cat needs to clear a 30 m high wall.
- Horizontal displacement: The cat needs to land approximately 250 m away from the cannon.
- Launch angle: The cat is launched at an angle of 55º.
- Launch height: The cat is assumed to be on the ground during launch.
To solve this problem, we can use the following equations of motion for projectile motion:
Vertical displacement (y-direction):
h = (v₀ * sin(θ))^2 / (2 * g)
Horizontal displacement (x-direction):
R = v₀ * cos(θ) * t
where:
- h is the vertical displacement (30 m),
- R is the horizontal displacement (250 m),
- v₀ is the launch speed,
- θ is the launch angle (55º),
- g is the acceleration due to gravity (approximately 9.8 m/s²),
- t is the time of flight.
Since we're given the vertical displacement, horizontal displacement, and launch angle, we can solve for the launch speed.
First, let's find the time of flight using the vertical displacement equation:
30 = (v₀ * sin(55º))^2 / (2 * 9.8)
Simplifying the equation:
15 * 9.8 = v₀^2 * sin²(55º)
v₀^2 = (15 * 9.8) / sin²(55º)
v₀^2 = 225 / sin²(55º)
v₀ ≈ √(225 / sin²(55º))
Next, let's find the launch speed using the horizontal displacement equation:
250 = v₀ * cos(55º) * t
Now, we need to find the time of flight (t). Rearranging the equation:
t = 250 / (v₀ * cos(55º))
Substituting the previously calculated value of v₀:
t = 250 / (√(225 / sin²(55º)) * cos(55º))
Finally, to convert the result into km/h, we'll multiply the launch speed by 3.6 since 1 m/s equals 3.6 km/h.
Launch speed (in km/h) = v₀ * 3.6
By calculating the above expressions, we can find the launch speed needed to make the cat clear the wall and land approximately 5 m over it.