The best leaper in the animal kingdom is the puma, which can jump to a height of 11.7 ft when leaving the ground at an angle of 51°. With what speed, in SI units, must the animal leave the ground to reach that height?
Y^2 = Yo^2 + 2g*h = 0
Yo^2 - 64*11.7 = 0
Yo^2 = 748.8
Yo = 27.4 Ft/s = Ver. component of
the initial velocity.
Vo = Yo/sinA = 27.4/sin51 = 35.2 Ft/s =
Initial velocity = Speed at which the
animal must leave gnd.
Vo = 35.2 Ft/s * 1m/3.3Ft = 10.7 m/s.
To determine the necessary speed for the puma to reach a height of 11.7 ft, we can use the principles of projectile motion. The height reached by a projectile is primarily determined by its initial vertical velocity.
Given:
- Height (h) = 11.7 ft
- Launch angle (θ) = 51°
To calculate the initial vertical velocity, we can use the following equation:
vₒ = √(2gh) / sinθ
Where:
- vₒ represents the initial vertical velocity
- g is the acceleration due to gravity (approximately 9.8 m/s²)
- h is the height in meters (to convert 11.7 ft to meters, multiply by 0.3048)
First, let's convert the height from feet to meters:
11.7 ft * 0.3048 m/ft = 3.56616 m (rounded to 5 decimal places)
Now we can calculate the initial vertical velocity (vₒ):
vₒ = √(2 * 9.8 m/s² * 3.56616 m) / sin(51°)
Using a calculator to perform the calculations, we get:
vₒ ≈ 7.260 m/s
Therefore, the puma must leave the ground with an initial vertical velocity of approximately 7.260 m/s in order to reach a height of 11.7 ft.