the puma, which can jump to a height of 12.9 ft when leaving the ground at an angle of 44°. With what speed, in SI units, must the animal leave the ground to reach that height?
The vertical component of velocity must be
Vf^2=2 g 12.9 where g=32ft/sec^2
Vf= sqrt(2*32*12.9)= 28ft/sec
so the velocity of the animal then is Vf/sin44 or about 28*1.4
you can do it more accurately.
To find the speed at which the puma must leave the ground to reach a height of 12.9 ft (converted to meters) at an angle of 44°, we need to use the principles of projectile motion.
First, let's convert the height from feet to meters.
1 ft = 0.3048 meters
12.9 ft = 12.9 * 0.3048 = 3.93252 meters (rounded to 5 decimal places)
Now, we need to break down the initial velocity into horizontal and vertical components. The vertical component is related to the height the puma reaches, while the horizontal component remains constant.
The vertical motion can be described using the equation:
y = u * sin(theta) * t - 0.5 * g * t^2
Where:
y = vertical distance (3.93252 meters)
u = initial velocity (unknown)
theta = angle of projection (44°)
t = time of flight (unknown)
g = acceleration due to gravity (9.8 m/s^2)
At the peak of its trajectory, the vertical displacement is zero. This happens when the puma reaches its maximum height.
Using this information, we can rewrite the equation as:
0 = u * sin(theta) * t - 0.5 * g * t^2
Next, we need to find the time of flight, t. At the maximum height, vertical velocity (v) is zero. We can use this information to find t.
v = u * sin(theta) - g * t
0 = u * sin(theta) - g * t
u * sin(theta) = g * t
t = u * sin(theta) / g
Substituting the value of t into the equation for vertical displacement:
0 = u * sin(theta) * (u * sin(theta) / g) - 0.5 * g * (u * sin(theta) / g)^2
0 = (u^2 * sin^2(theta)) / g - 0.5 * u^2 * sin^2(theta) / g
0 = (u^2 * sin^2(theta) - 0.5 * u^2 * sin^2(theta)) / g
Simplifying the equation:
0.5 * u^2 * sin^2(theta) = 0
This equation tells us that either u = 0 or sin^2(theta) = 0.
Since the puma is leaving the ground with some velocity, u cannot be zero. Therefore, we take the alternative, sin^2(theta) = 0.
sin^2(theta) = 0
sin(theta) = 0
Now, we know that θ = 44° and sin(44°) = 0.69465.
sin(theta) = 0.69465
0.5 * u^2 * (0.69465)^2 = 0
Solving for the initial velocity, u:
u^2 = 0 / (0.5 * (0.69465)^2)
u^2 = 0
Therefore, the initial velocity, u, is equal to 0 m/s according to this analysis.
In conclusion, for the puma to reach a height of 12.9 ft with an angle of 44 degrees, it would need to leave the ground with an initial velocity of 0 m/s.