Two children are playing on two trampolines. The first child can bounce 1.4 times higher than the second child. The initial speed up of the second child is 4.8 m/s. The maximum height of the second child is 1.17m and the initial speed of the first child is 5.66 m/s. How long was the first child in the air?
To find the time the first child was in the air, we can use the kinematic equation for vertical motion:
h = v_i * t + (1/2) * a * t^2,
where
h is the height,
v_i is the initial velocity,
t is the time,
and a is the acceleration.
Let's find the time for the second child first.
Given:
For the second child,
v_i2 = 4.8 m/s (initial speed),
h2_max = 1.17 m (maximum height).
We know that the maximum height is reached when the vertical velocity of the child becomes 0.
Using the equation v_f2 = v_i2 + a2 * t, where v_f2 is the final velocity (0 in this case), we can solve for the acceleration:
0 = 4.8 m/s + a2 * t.
Since the child only goes up and comes back down, the time it takes to reach the maximum height is half of the total time in the air. Therefore, we can rewrite the equation as:
0 = 4.8 m/s + (1/2) * a2 * (t/2),
Simplifying,
0 = 4.8 m/s + 0.5 * a2 * (t/2)
0 = 4.8 m/s + 0.25 * a2 * t,
Since the child comes back down with the same acceleration, we can rewrite the equation as:
0 = 4.8 m/s - 0.25 * a2 * t,
Rearranging,
a2 = (4.8 m/s) / (0.25 * t).
Now let's find the acceleration of the second child (a2):
a2 = (4.8 m/s) / (0.25 * t).
We know that the initial speed of the second child is 4.8 m/s, so acceleration can be calculated as:
a2 = (4.8 m/s) / (0.25 * t) = 19.2 / t m/s^2.
Given the maximum height of the second child is 1.17 m, we can find the time taken to reach this height using the kinematic equation:
h2_max = v_i2 * t + (1/2) * a2 * t^2.
Substituting the known values,
1.17 m = 4.8 m/s * t + (1/2) * (19.2 / t) * t^2.
1.17 = 4.8t + 19.2t.
Now we can solve this quadratic equation to find the value of t.