Solving SUVAT Questions: Find Time for a Ball's Descent

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SUMMARY

The discussion centers on solving SUVAT equations to determine the time it takes for a ball, launched vertically at 10 m/s, to hit the ground. The user initially calculated the time using the formula v = u + at, resulting in an incorrect time of 1.02 seconds. The correct approach involves using the equation d = Vo*t + 0.5 * a * t^2, where d equals 0, Vo is 10 m/s, and a is -9.81 m/s². The user is advised to consider the motion in two phases: ascent and descent, to accurately compute the total time.

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jendrix
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Hi, I'm just practicing some SUVAT style questions as part of my revision and have a quick question.

Here's an example, a ball is catapulted vertically upwards at 10m/s, find the time it takes to hit the ground.(Discount air resistance)

I used S =0 U=10 m/s V=? A=-9.8 T=?

Using v =u +at gave me a time of 1.02s however this is wrong.

Is this because I've done this as one motion? Should I split it into to, ie find highest position then journey back down?

Thanks

Just to reiterate this isn't a homework question.
 
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The ball goes up and the ball returns to the ground. So double the time it takes to attain a velocity of 10 m/sec if dropped.

Or solve the following:

d = Vo*t + .5 * a * t^2

where d=0.0, Vo=10. m/sec, and a = -9.81 m/sec^2
 
jendrix said:
I used S =0 U=10 m/s V=? A=-9.8 T=?

Using v =u +at gave me a time of 1.02s however this is wrong.

What is value of V?
 

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