- #1
jacksonbobby5
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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?
How Far Does a Kangaroo Jump on Planet Y?
- Thread starter jacksonbobby5
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SUMMARY
The discussion focuses on calculating the jump distance of a kangaroo on Planet Y, where gravity is 12 m/s² and the initial jump velocity is 8 m/s. Participants suggest using kinematic equations, specifically y = (Vi²) / (2g), to determine the jump height. The total time of flight is calculated to be 1.333 seconds, with each half of the jump (ascent and descent) taking 0.666 seconds. The conversation emphasizes the need for the angle of the jump to accurately compute the horizontal distance traveled.
PREREQUISITES- Understanding of kinematic equations
- Knowledge of gravitational acceleration
- Familiarity with initial velocity concepts
- Basic principles of projectile motion
- Research the kinematic equation y = (Vi²) / (2g) for projectile motion
- Learn about calculating time of flight in projectile motion scenarios
- Explore the effects of angle on projectile distance
- Study the impact of atmospheric resistance on projectile motion
Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion dynamics.
- #2
Hootenanny
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What have you attempted thus far? What are your thoughts?
- #3
jacksonbobby5
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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.
- #4
Hootenanny
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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.jacksonbobby5 said:
- #5
jacksonbobby5
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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?
- #6
Hootenanny
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Sounds goot to mejacksonbobby5 said:
- #7
misnoma
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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?
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?