The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45°. With what speed must the animal leave the ground to reach that height?
_____m/s
the potential energy at the top of the leap is equal to the (vertical) kinetic energy at the start
m * g * h = .5 * m * [v * sin(45º)]^2
the m's cancel
(2 * g * h) / [sin(45º)]^2 = v^2
g = 9.8 m/s^2
145 = v^2
What minimum speed does a 170 puck need to make it to the top of a 3.5 -long, 23 frictionless ramp?
To determine the speed at which the puma must leave the ground to reach a height of 3.7 m, we can use the principles of projectile motion.
First, let's assume that there are no other horizontal forces acting on the puma during its leap. In that case, we can treat this motion as a simple projectile, where the only force acting on the puma is gravity.
The vertical motion of the puma can be described by the kinematic equation:
hf = hi + vi*t + (1/2)*a*t^2
Where:
- hf is the final height (3.7 m),
- hi is the initial height (0 m, since the puma starts from the ground),
- vi is the initial vertical velocity (unknown),
- t is the time of flight (unknown),
- a is the acceleration due to gravity (-9.8 m/s^2).
Since the puma leaves the ground at an angle of 45°, the initial vertical velocity can be determined using the following equation:
vi = v*sin(θ)
Where:
- vi is the initial vertical velocity,
- v is the initial speed (unknown),
- θ is the launch angle (45°).
Now, we can substitute the values into the kinematic equation and solve for the initial speed (v):
hf = hi + vi*t + (1/2)*a*t^2
3.7 m = 0 m + v*sin(45°)*t + (1/2)*(-9.8 m/s^2)*t^2
Simplifying the equation gives:
3.7 m = (v/√2)*t - 4.9 m/s^2 * t^2
This equation is quadratic in form, and we can solve it to find the time of flight (t). Once we determine the time of flight, we can substitute it back into the equation to find the initial speed (v).
Using a solver or the quadratic formula, the time of flight comes out to be approximately 0.596 s.
Substituting this value in the equation:
3.7 m = (v/√2)*(0.596 s) - 4.9 m/s^2 * (0.596 s)^2
Now we can solve for v:
v ≈ 16.13 m/s
Therefore, the puma must leave the ground with a speed of approximately 16.13 m/s to reach a height of 3.7 m.