- #1

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say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.

so cant use s=v*t

does anyone kow What is the equation for this?

- Thread starter ghostbuster25
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- #1

- 102

- 0

say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.

so cant use s=v*t

does anyone kow What is the equation for this?

- #2

Kurdt

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https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Find one without time involved.

- #3

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You must know these concept. All the vectors you must consider 2 components. Horizontal Motion velocity is constant and vertical motion acceleration is constant.If the particle dropped from the hill, no need to consider angle for initial velocity.

say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.

so cant use s=v*t

does anyone kow What is the equation for this?

Horizontal Motion Vertical motion

initial velocity v(0x)= v(0) v(0y)=0 (for above case)

final velocity v(x)=v(0x)=v(0) v(y)=.........

acceleration a(x)=0 a(y)=-g

displacement x y

time t t (scalar)

x=v(0x)t y=v(0y)t -0.5 gt^2

v(y)=v(0y) -gt

v^2=v(0y)^2-2gy

You may use simple free fall bodies equations.

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