
#1
Mar2311, 02:28 PM

P: 3

1. The problem statement, all variables and given/known data
In projectile motion, the time it takes for an object to get from a certain height to the peak is the same time as the time to reach the same height from the peak, correct? Well then, I came across something in which an object was shot in the air via a slingshot and peaked at about 1 second. Then, it took approximately 2 seconds to return to the ground at the same height. Is this possible? If so, how? (Lets say that there may be air resistance and wind as an influence) 2. Relevant equations Kinematics Equations: V = Vo +at x = 1/2(Vo+V)t x = Vit + 1/2at^2 V^2 = Vo^2 + 2ax 3. The attempt at a solution I thought that when in projectile motion, an object would have the same time reaching the peak as it would reaching the ground from the peak. (Symmetry). 



#2
Mar2311, 02:52 PM

P: 47





#3
Mar2311, 03:11 PM

P: 3





#4
Mar2311, 03:16 PM

P: 4

Projectile Motion Time Question
can someone help me with the post need help...please....




#5
Mar2311, 03:43 PM

P: 47

Drag  much like friction  slows an object down, so if the object has an x and y component of velocity, it would reduce both. In a similar way, if you drew x and y axis on a table top, then slid a note card across table, the notecard's x and y velocity would slow down with friction. But the drag force also depends on the speed (both x and y), so it's a bit harder to work with in two dimensions. The get an idea of how it works, consider shooting an object straight up, so we only have to work on one dimension. We'll first consider the fall from the maximum height. The object accelerates downward because of gravity, but as it starts moving faster, the drag increases, providing an upward force. The end result is that the object approaches its terminal velocity, and gets very close, but doesn't go any faster. Not since you can shoot the object faster than the terminal velocity, you can get it up to the highest point faster than it comes down. 


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