If a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second, for how long will the ball be going upward?

Answer 1
The answer is: #1,73s#.
First of all: #v=56(feet)/s~=17m/s#.

This is an uniformed accelerated motion with

#a=-g=-9.8m/s^2#, and the law is:
#v=v_0+atrArrv=v_0-g trArrt=(v_0-v)g#.
Since #v_0=17m/s# and #v=0m/s# (in the higher point his speed is 0),
#t=(17m/s)/(9.8m/s^2)=1,73s#.
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Answer 2

To find the time the ball will be going upward, you can use the equation for vertical motion:

[ v_f = v_i - gt ]

Where: ( v_f ) = final velocity (when the ball reaches its peak, its final velocity is 0) ( v_i ) = initial velocity (given as 56 feet per second) ( g ) = acceleration due to gravity (approximately -32 feet per second squared, negative because it acts downward) ( t ) = time

Rearranging the equation to solve for ( t ):

[ t = \frac{{v_f - v_i}}{{-g}} ]

Plug in the values:

[ t = \frac{{0 - 56}}{{-32}} ]

[ t = \frac{{-56}}{{-32}} ]

[ t = 1.75 \text{ seconds} ]

So, the ball will be going upward for 1.75 seconds.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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