[SOLVED] Just Checking 1. The problem statement, all variables and given/known data In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80 . A flea jumps straight up to a maximum height of 0.550m . What is its initial velocity Vsub0 as it leaves the ground? Express your answer in meters per second to three significant figures. 2. Relevant equations Since my givens are x = .550m [Vsub0] = unknown [Asubx] = =9.80 Hopefully [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 3. The attempt at a solution [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in 0^2 = [Vsub0x]^2 + 2[-9.80]*(.550) 0 = [Vsub0x]^2 -10.78 10.78 = [Vsub0x]^2 Sqrt(10.78) = 3.28 m/s I think this is correct but i want to check as the homework system is somewhat unforgiving
Now that I completed that the homework is asking me another question. 1. The problem statement, all variables and given/known data How long is the flea in the air from the time it jumps to the time it hits the ground? Express your answer in seconds to three significant figures. x = .550m [Vsub0] = 3.28m/s (ans to 1) [Asubx] = =9.80m/s^2 t= unknown 2. Relevant equations X-[Xsub0] ={ ([Vsub0x] + [Vsubx])/2 } t 3. The attempt at a solution x - xsub0 is .550m because initial position is 0 and final is .550m .550m ={ ([Vsub0x] + [Vsubx])/2 } t sub in vars .550m ={ ([3.28m/s] + [0])/2 } t .550m ={ 3.28/2 } t .550m = 1.64m/s * t .550 / 1.64 = .335 ans = .335 to go from gound to top of parabola .670 to go all the way? Just checking again, this new to me.
Use time of fall = time of rise. Makes things easier. h = 1/2 gt^2 => 0.55 = 0.5*9.8*t^2 => t = 0.33 => 2t = 6.7.