Finding the speed of a ball given some height and falling speed

AI Thread Summary
Nicole throws a ball upward, and Chad observes it passing his window at a height of 6.10 m with a speed of 11.0 m/s on its way down. The key to solving the problem involves understanding that the ball's speed decreases due to gravity as it ascends and increases as it descends. To find the initial speed at which Nicole threw the ball, the equation v^2 = u^2 + 2as can be used, where v is the final velocity (11 m/s), a is the acceleration due to gravity (-9.8 m/s²), and s is the displacement (6.1 m). By rearranging the equation and solving for the initial speed, one can determine how fast Nicole threw the ball.
skysunsand
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Homework Statement



Nicole throws a ball straight up. Chad watches the ball from a window 6.10 m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 11.0 m/s as it passes him on the way back down. How fast did Nicole throw the ball?

Homework Equations



x= x0 + v0t + 1/2 at^2
v= v0+at

The Attempt at a Solution



I haven't the slightest clue as how to solve this. If she's tossing it upward, I know that would be a parabola graphically, so perhaps I need to use the quadratic equation at some point.
Need to find acceleration, I suppose, to find how fast she threw the ball?
 
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skysunsand said:

Homework Statement



Nicole throws a ball straight up. Chad watches the ball from a window 6.10 m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 11.0 m/s as it passes him on the way back down. How fast did Nicole throw the ball?

Homework Equations



x= x0 + v0t + 1/2 at^2
v= v0+at

The Attempt at a Solution



I haven't the slightest clue as how to solve this. If she's tossing it upward, I know that would be a parabola graphically, so perhaps I need to use the quadratic equation at some point.
Need to find acceleration, I suppose, to find how fast she threw the ball?

The acceleration involved is the acceleration due to gravity.
If the ball was doing 11 m/s on its way down past the window, it will have been traveling at 11 m/s on its way up past the window as well.

Be careful about which figures are positive and negative when calculating.
 
So...the answer is just 11 then? It was a trick question?
 
skysunsand said:
So...the answer is just 11 then? It was a trick question?

NO, it had slowed to 11 m/s by the time it went past the window, it must have been going faster when it left her hand.

Peter
 
So it slows down from some unknown number due to gravity. Which would mean negative acceleration. But then how do you figure out the unknown, and how does 11 work into the equation?
 
skysunsand said:
So it slows down from some unknown number due to gravity. Which would mean negative acceleration. But then how do you figure out the unknown, and how does 11 work into the equation?

When I do these problems I use the following symbols

v = final velocity
u = initial velocity
s = displacement
a = acceleration
t = time

For the ball, we know it had slowed to 11 m/s by the time it got to the window 6.1 m up

so
v = 11
a = -9.8
s = 6.1

You want to know the initial speed , "u" , so the formula of choice would be

v^2 = u^2 + 2as [It is the only one of the 5 that does not require a time value]

121 = u^2 + 2*-9.8*6.1

you should be able to do it from there. Be careful with that minus sign.
 
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