The cool thing to consider here is the fact that you can break the motion of the ball into three parts

from #"50 m"# above the ground to max height #-># going upfrom max height to #"50 m"# above the ground #-># going downfrom #"50 m"# above the ground #-># going downSo if you throw the ball from a height of #"50 m"# with an initial velocity of #"30 m s"^(-1)#, it will *decelerate* under the influence of gravity until it comes to a complete stop at the top of its trajectory, i.e. at maximum height, then *accelerate* under the influence of gravity until it reaches the ground.

So, if the velocity of the ball is equal to #"0 m s"^(-1)# at the top of its trajectory, you can say that you have

#"0 m s"^(-1) = v_0 - g * t_ "up from 50 m"#

which, in your case, is equal to

#"0 m s"^(-1) = "30 m s"^(-1) - "9.81 m s"^(-2) * t_ "up from 50 m"#

This means that the ball will reach the peak of its trajectory in

#t_ "up from 50 m" = (30 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.81color(red)(cancel(color(black)("m"))) "s"^color(red)(cancel(color(black)(-2)))) = "3.06 s"#

Now, when the ball reaches #"50 m"# above the ground **on its way down**, its velocity will actually be **equal** to the initial velocity, only it will be pointed towards the ground.

This happens because the ball is in *free fall* from its maximum height, so the time it takes for the ball to reach #"50 m"# **on its way down** will be equal to the time it took for it to reach maximum height.

#"time up from 50 m to max height = time down from max height to 50 m"#

So

#t_"down to 50 m" = "3.06 s"#

So you know for a fact that when the ball reaches #"50 m"# above the ground **going down**, it will have a velocity of #"30 m s"^(-1)#.

This means that for the last portion of the movement, you can look consider a ball thrown from #"50 m"# with #"30 m s"^(-1)# **pointed down**.

You are watching: A ball is thrown up from the top of the tower

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You can find the impact velocity by using the equation

#v_f^2 = v_0^2 + 2 * g * h#

This will get you

#v_f = sqrt(("30 m s"^(-1))^2 + 2 * "9.8 m s"^(-2) * "50 m")#

#v_f = "43.37 m s"^(-1)#

This means that you have

#v_f = v_o + g * t_ "50 m to ground"#

which gets you

#t_"50 m to ground" = (43.37 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))) - 30 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.81color(red)(cancel(color(black)("m"))) "s"^color(red)(cancel(color(black)(-2))))#

#t_"50 m to ground" = "1.36 s"#

Therefore, you can say that the **total time** needed for the ball to hit the ground is

#t_"total" = t_"up from 50 m" + t_ "down to 50 m" + t_ "50 m to ground"#

#t_ "total" = "3.06 s" + "3.06 s" + "1.36 s"#

#color(darkgreen)(ul(color(black)(t_"total" = "7.5 s")))#

I"ll leave the answer rounded to two **sig figs**, but keep in mind that you only have one significant figure for your values.