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

AI Thread Summary
A ball is thrown upward with an initial velocity of 20 m/s, remaining in the air for a total of 4.08 seconds and reaching a maximum height of 20.4 meters. To determine when the ball is 15 meters above the ground, the equation y(t) = v₀t - 1/2gt² can be used, where y is set to 15m. Users are encouraged to solve for time 't' using this equation. Additional resources, such as a simulation for projectile motion, are suggested for further understanding. The discussion emphasizes the importance of correctly applying the kinematic equations to solve for specific heights.
carltonblues
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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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