a kangaroo jumps to a vertical height of 2.55m.How long was it in the air before returning to Earth?
pleaseee anwser it is due in 1 day!!!!
How long does it take to fall from 2.55 m ?
2.55 = (1/2)9.8 t^2
t^2 = .520
t = .141s
twice that so it can go up = .282 seconds
THANK YOU FOR YOUR HELP!!!
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To find the time the kangaroo was in the air before returning to Earth, we can use the law of physics that states:
"the time taken for an object to fall vertically from a certain height is equal to the time it took to reach that height when it was thrown upwards."
We can use this law to calculate the time it took for the kangaroo to reach its maximum height. The formula for the time taken to reach the maximum height (t_up) is:
t_up = v / g
where:
v = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
Since we know the vertical velocity at the highest point is 0 (as the kangaroo is momentarily at rest), we can assume the initial vertical velocity is equal to the final vertical velocity at the highest point.
Next, we need to find the initial vertical velocity (u) using the formula:
v = u + gt
where:
v = final vertical velocity (0 m/s)
u = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
t = time taken to reach the maximum height
Since v is 0, the formula simplifies to:
0 = u + gt_up
Solving for u, we get:
u = -gt_up
Now, we can substitute this value of u into the first equation to find t_up:
t_up = u / g
t_up = (-gt_up) / g
t_up = -t_up
We have obtained the value of t_up, which represents the time the kangaroo took to reach its maximum height. Now, we need to double this value to find the total time the kangaroo was in the air before returning to Earth:
t_total = 2 * t_up
Let's calculate the value of t_total:
t_up = 2.55 m / 9.8 m/s^2
t_total = 2 * t_up
Calculating t_total, we find:
t_total ≈ 0.52 seconds
Therefore, the kangaroo was in the air for approximately 0.52 seconds before returning to Earth.