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?
we don't know for sure that the max height is 35m thought, so oobleck's answer might not actually be accurate.
To find the speed at which the cat needs to be launched, we can use the principles of projectile motion.
First, we need to break down the initial velocity of the cat into its horizontal and vertical components.
The initial velocity in the horizontal direction (Vx) remains constant throughout the motion and can be determined using the equation:
Vx = V * cos(θ)
Where V is the initial velocity of the cat and θ is the launch angle (55º).
The initial velocity in the vertical direction (Vy) can be determined using the equation:
Vy = V * sin(θ)
Since the cat needs to clear a 30 m high wall, we can set up the following equation to determine the time it takes for the cat to reach its maximum height (h):
h = Vy * t - (1/2) * g * t^2
Where g is the acceleration due to gravity (approximately -9.8 m/s^2) and t is the time.
To find the total time of flight, we can assume that it takes the same amount of time to reach the maximum height and fall back down. Therefore, the total time of flight is twice the time it takes to reach the maximum height:
t_total = 2 * t
Now, we can determine the horizontal distance traveled (d) by the cat using the equation:
d = Vx * t_total
Given that the wall is 250 m from the cannon and the cat needs to clear it by approximately 5 m, we have the following equation:
d = 250 m
h + 5 = 30 m
Now, we can solve these equations simultaneously to find the initial velocity (V) in m/s.
1. Rearrange the equation for d:
250 m = V * cos(θ) * 2 * t
2. Rearrange the equation for h:
25 m = V * sin(θ) * t - (1/2) * g * t^2
3. Substitute for t_total in the equation for d using t_total = 2 * t:
250 m = V * cos(θ) * 2 * (1/2) * t_total
4. Rearrange the equation for h to isolate t^2:
25 m = V * sin(θ) * t - (1/2) * g * t^2
0 = - (1/2) * g * t^2 + V * sin(θ) * t - 25 m
5. Substitute for t from step 3 into the equation from step 4:
0 = - (1/2) * g * (t_total/2)^2 + V * sin(θ) * t_total/2 - 25 m
6. Rearrange the equation from step 5 to isolate V:
0 = - (1/2) * g * (t_total/2)^2 + (V * sin(θ))/2 * t_total - 25 m
7. Substitute the known values into the equation from step 6:
0 = - (1/2) * (-9.8 m/s^2) * (t_total/2)^2 + (V * sin(55º))/2 * t_total - 25 m
8. Solve the equation from step 7 for V in m/s.
To solve this problem, we can use the principles of projectile motion. The motion of the cat can be divided into horizontal and vertical components.
First, let's find the initial vertical velocity component (Vy) of the cat. We know that the vertical displacement (Δy) is 5 m and the acceleration due to gravity (g) is approximately 9.8 m/s². Since the cat is launched at an angle of 55º from the ground, we can use trigonometry to find the initial vertical velocity:
Vy = Δy / (t + 1/2 * g * t^2)
where t is the time it takes for the cat to reach the wall.
Next, let's find the horizontal component of the cat's velocity (Vx). We know that the horizontal displacement (Δx) is 250 m and the time it takes for the cat to reach the wall is the same as the time of flight (t). So:
Δx = Vx * t
Now, we can find the total velocity of the cat (V) using the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)
Finally, to convert this velocity to km/h, we can multiply by 3.6:
V(km/h) = V(m/s) * 3.6
By plugging in the given values into these equations, we can find the velocity at which the cat needs to be launched.
Note: We assume there is no air resistance in this problem, and all calculations are approximate.
I hope this explanation helps!
just solve
v cos55º t = 250
v sin55º t - 4.9t^2 = 35