Optimum Angle for Jumping: Height & Distance

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The optimum angle for achieving the greatest height in a jump is typically 90 degrees, while the angle for maximum distance is generally 45 degrees on level ground. When launching from a surface below the landing surface, the optimum angle is greater than 45 degrees to compensate for the height difference. The discussion highlights the importance of understanding projectile motion equations to calculate the height and distance of jumps. Additionally, specific examples, such as calculating the power needed for a BMX jump, illustrate the practical application of these concepts. Understanding these angles and formulas is crucial for optimizing performance in jumping activities.
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Homework Statement


What is the optimum angle for greatest height? What is the greatest optimum for greatest distance? When on a surface below the landing surfaces is the optimum angle less than or greater than 45 degrees? What is the formula for finding the distance/height of the angle?

example question:
if the BMX bike wants to jump on the ledge that is 3.5ft and 7ft away what is the amount of power he needs to use? What is the optimum angle of the jump?
2. Relevant equ


3. The Attempt at a Solution [/b]
I think the optimum angle for a landing surface above the starting surface is greater than 45 degrees. I have no idea what the formula is though.
 
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In projectile motion, the greatest height is achieved when the object is projected directly upwards. For other things just find the equation of a projectile motion.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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