A rock is tossed upwards at 15.0 m/s from a height of 1.00m above the ground, what will be its velocity when it is 6.00m above the ground?

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Explanation:

You know that your ball is being thrown from a height of 3.6 m with an initial velocity of 2.8 m/s.

At the top of its motion, i.e. at maximum height, the velocity of the ball will be zero. This means that you can determine the height it travels before starting its decent by using the initial velocity

v

2

top



=0

=

v

2

i

−

2

⋅

g

⋅

h

up

This is equivalent to

v

2

i

=

2

⋅

g

⋅

h

up

⇒

h

up

=

v

2

i

2

⋅

g

h

up

=

2.8

2

m

2

⋅

s

−

2

2

⋅

9.8

m

⋅

s

−

2

=

0.40 m

The ball will be at a total height of

h

=

h

i

+

h

up

=

3.6

+

0.40

=

4.0 m

Now use the same equation again, only this time take into account the fact that the initial velocity, i.e. the velocity at maximum height, is zero.

v

2

f

=

v

2

top



=0

+

2

⋅

g

⋅

h

v

f

=

√

2

⋅

g

⋅

h

=

√

2

⋅

9.8

m

⋅

s

−

2

⋅

4.0

m

v

f

=

8.9 m/s

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