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?
answer in (m/s)
12.04
Initial velocity in the y direction must be.
v^2= 2 (9.8) 3.7 or
v = sqrt(2*9.8*3.7)
Divide this by sin45 to get the total init vel
12m
To find 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, we need to understand that the height reached by the puma depends on the vertical component of its velocity. We can calculate this component using the equation:
v_vertical = v_initial * sin(θ)
Where:
- v_vertical is the vertical component of velocity
- v_initial is the initial velocity of the puma
- θ is the launch angle (45° in this case)
Given that the puma reaches a height of 3.7 m, we can set v_vertical equal to the final vertical velocity when the puma reaches its maximum height:
v_vertical = 0 m/s
So we have:
0 = v_initial * sin(45°)
To isolate v_initial, we divide both sides of the equation by sin(45°):
v_initial = 0 / sin(45°)
Using a calculator, we find that sin(45°) is equal to √2 / 2 or approximately 0.7071.
Therefore:
v_initial = 0 / 0.7071 = 0
This means that the puma must leave the ground with a minimum speed of 0 m/s to reach a height of 3.7 m. However, it's important to note that this answer is not physically possible. In reality, the puma would need some velocity to reach the given height.