How would be the number 54 be correctly produced in ASL
2 answers:
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The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s
Why?
To solve the problem, we must consider the vertical height reached by the student as max height.
We can use the following equations to solve the problem:
<u>Initial speed calculations:</u>
At max height, the speed tends to zero.
So, calculating, we have:
<u></u>
<u>Hang time calculations:</u>
We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:
Then, for the total hang time, we have:
Have a nice day!
G = 9.81 m/sec^2) g = 9.81
<span>Solving for velocity : </span>
<span> = 2gh </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
<span>Solving for velocity : </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
Answer:
b 1.39 m/s²
Explanation:
Given the following data;
Time = 12 seconds
Distance, S = 100 m
Since it's starting from rest, the initial velocity is equal to 0m/s.
To find the acceleration, we would use the second equation of motion;
Where;
S represents the displacement or height measured in meters.
u represents the initial velocity measured in meters per seconds.
t represents the time measured in seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
100 = 0(12) + ½*a*12²
100 = 0 + 72
100 = 72a
Acceleration, a = 100/72
Acceleration, a = 1.389 ≈ 1.39 m/s²