Best Answer

The ball reaches a maximum height of ~81.6 meters.

How to do it:

Find the change in time using Vf = Vi -g(t)

We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question)

So, Vf = Vi - g(t) => 0 = 40 - g(t) --Rework the equation to solve for t...

t = 40/g => since g = 9.8... t = 40/9.8 ~ 4.08

So the change in time is about 4.08 seconds

Now plug your new values into the kinematics equation

Yf = Yi + Vi(t) + 1/2(a)(t^2)

Plug in our values and solve => Yf = 0 + (40 * 4.08) - (9.8 * 4.08^2)/2

Therefore, Yf = 81.6 meters. This is your maximum height. Remember to find the time taken for the ball to fall from this height back to the ground if you need to calculate the total time of the fall.

Source: I recently took a test with this exact question, got it right.

Thanks for posting!

~Secret Physics Man

Q: A ball is thrown straight up with an initial speed of 40 ms How high does it go?

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Its initial speed cannot be 20 m, as stated in the question. Secondly, if the initial speed is correctly given, then there is no need to calculate it!

The initial velocity is zero. In most basic physics problems like this one the initial velocity will be zero as a rule of thumb: the initial velocity is always zero, unless otherwise stated, or this is what you are solving for Cases where the initial velocity is not zero examples a cannon ball is shot out of a cannon at 50 mph a ball is thrown from at a speed of 15 mph etc

That depends on whether you mean a golf ball, or something large and light like a volleyball,which is more venerable to air resistance, and would have to have a higher initial speed to reach 60m.

... and what is the question? The second ball should arrive at the floor a second after the first, both should have the same speed.

They should reach the ground together, since their initial vertical speed is the same, namely zero.

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Its initial speed cannot be 20 m, as stated in the question. Secondly, if the initial speed is correctly given, then there is no need to calculate it!

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