How Far Does a Kangaroo Jump on Planet Y?

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To determine how far a kangaroo jumps on planet Y with a gravity of 12 m/s² and an initial velocity of 8 m/s, the discussion revolves around using kinematic equations. The formula y = (Vi^2)/(2g) is suggested as a potential solution, although its validity is questioned. The time of flight for the jump is calculated to be 1.333 seconds, with 0.666 seconds for both ascent and descent. The conversation emphasizes the need for the angle of the jump to calculate the horizontal distance traveled. Overall, the discussion focuses on applying kinematic principles to solve the problem of the kangaroo's jump distance.
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So if I have a kangaroo on planet Y with a specified gravity of 12m/s^2. If he jumps with an inital velocity of 8m/s, how far would he travel?
 
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What have you attempted thus far? What are your thoughts?
 
Well I thought I would use the formula y=Vi(time)+1/2g(time)^2, but I don't have a time, just initial velocity and gravity.
 
jacksonbobby5 said:
Well I thought I would use the formula y=Vi(time)+1/2g(time)^2, but I don't have a time, just initial velocity and gravity.
Which other kinematic equations do you know? You need to pick one with all the variables in that you know plus the one which you want to find out.
 
Well, I also had read about an equation that is y=(Vi^2)/2g but I wasnt sure wether this was a valid equation or a kinematic equation. Any clue?
 
jacksonbobby5 said:
Well, I also had read about an equation that is y=(Vi^2)/2g but I wasnt sure wether this was a valid equation or a kinematic equation. Any clue?
Sounds goot to me :approve:
 
assume a level ground with no atmospheric resistance.

time of flight for the first half (rising) is t = (v-u)/g = 0.666 sec, sot he time of flight for the last half descent is also 0.666 sec. Total time is thus 1.333 sec.

with no reistance to motion in the x-axis the distance traveled is just the (X component of velocity) x time, do you know the angle?
 
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