The ball reaches the ground after 7.76 s
Step-by-step explanation:
The motion of the ball along the vertical direction is a free fall motion, which is affected by the force of gravity only; therefore it is a uniformly accelerated motion (=constant acceleration), so we can use the following suvat equation:
[tex]s=ut-\frac{1}{2}gt^2[/tex]
where
s is the displacement
u is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
t is the time
In this problem,
u = 38 m/s
Also, we want to find the time t at which the ball hits the ground again: so, when the displacement becomes zero again,
s = 0
Therefore the equation becomes:
[tex]0=38t-\frac{1}{2}9.8t^2\\0=38t-4.9t^2[/tex]
And solving for t,
[tex]t(38-4.9t)=0[/tex]
we have two solutions:
t = 0 (instant at which the ball is kicked)
[tex]38-4.9t=0\\t=\frac{38}{4.9}=7.76 s[/tex]
which is the solution to the problem.
Learn more about free fall motion:
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