What is the time when the ball is 15 m above the ground?

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SUMMARY

The discussion focuses on calculating the time a ball is 15 meters above the ground after being thrown upward with an initial velocity of 20 m/s. The total time the ball is in the air is established as 4.08 seconds, with a maximum height of 20.4 meters. To find the time when the ball reaches 15 meters, the equation of motion y(t) = v0t - (1/2)gt2 is utilized, where g is the acceleration due to gravity (approximately 9.81 m/s2). Participants confirm that setting y to 15 meters and solving for t is the correct approach.

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1)A ball is thrown upward with an initial velocity of v = 20 m/s. How long is the ball in the air? What is the greatest height of the ball? When is the ball 15 m
above the ground?


1) I have figured out

t total = 4.08s
greatest height = 20.4m

For the last bit (Finding the time when x=15m), I am having trouble. I know the answer but cannot work it through.

Do I use s = ut+1/2at^2 for it?

If I do, it won't work out because the 20.4 m has to come into it, doesn't it?

Please help!
 
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The equation giving height (let's label it through 'y') as function of time is
y(t)=v_{0}t-\frac{1}{2}gt^{2}

Now set "y" to 15m and solve for "t"...

Daniel.
 
dextercioby said:
The equation giving height (let's label it through 'y') as function of time is
y(t)=v_{0}t-\frac{1}{2}gt^{2}

Now set "y" to 15m and solve for "t"...

Daniel.
Cheers mate
 

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