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Sagot :
Here we need to study the motion equations of an object.
We will see that:
The time that passes between the moment Gary throws the ball and the moment Alex catches it is t = 1.37 seconds.
Let's write the motion equations of the ball:
Any object that is on the air (when we ignore the air resistance and such) is only being accelerated by Earth's gravitational force.
Thus, the acceleration of the ball will be:
a(t) = -9.8m/s^2
Where the negative sign is because this acceleration is downwards.
Now to get the velocity of the ball we need to integrate over time, we will get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the initial velocity of the ball, which we know is 10m/s, then we have:
v(t) = (-9.8m/s^2)*t + 10m/s
To get the position equation of the ball we integrate again:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10m/s)*t + p0
where p0 is the initial position of the ball, we do not know it, and really we do not need it, so we can define p0 = 0m, then the position equation is:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10m/s)*t
Such that p(0s) = 0m
We know that the ball is caught when it is 4.5m above the position where it was thrown, so we need to solve:
p(t) = 4.5m
for t.
4.5m = (1/2)*(-9.8m/s^2)*t^2 + (10m/s)*t
(4.9 m/s^2)*t^2 - (10 m/s)*t + 4.5m = 0
This is a quadratic equation, so we can just use the Bhaskara's formula to get the solutions:
[tex]t = \frac{-(-10m/s) \pm \sqrt{(-10m/s)^2 - 4*(4,9m/s^2)*(4.5m)} }{2*(4.5 m/s^2)} \\\\t = \frac{(10m/s) \pm 3.44 m/s }{9.8 m/s^2}[/tex]
Then we have two solutions:
[tex]t = \frac{(10m/s) - 3.44 m/s }{9.8 m/s^2} = 0.67 s\\\\t = \frac{(10m/s) + 3.44 m/s }{9.8 m/s^2} = 1.37 s[/tex]
The first one is for the first time that the ball reaches the height of 4.5m (when it is going up) while the second one is for the part when the ball is falling down.
So the time that passes between the moment Gary throws the ball and the moment Alex catches it is t = 1.37 seconds.
If you want to learn more, you can read:
https://brainly.com/question/10674495
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