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a) With what initial speed did the man throw the ball?

Ans. H = 40m

t = 1.8 seconds

Using,

x-x0 = v0t + 1/2 at^2

To find vo:

x -xo = 0 ( I wonder why x -xo , is zero,) when the height of the cliff is given to be 40m.

0 = 1.8u + 4.9 * (1.8)^2

Solving for u, we get:

v = 8.82 m/s, which is correct I think!

b) How high above the point of release did the ball go?

We can use,

v^2 = v0^2 + 2g (x-xo)

v = final velocity = 0 m/s

v0 from part a, is 8.82 m/s

and g = 9.81 m/s2

Plugging in all that, we get:

x-x0 = 3.964 m

c) What is the speed of the ball when it passes him on the way back down?

Same as a) 8.82 m/s

d) How long after the man throws it does the ball hit the ground at the bottom of the cliff?

This is where I am stuck......

Which equation to use?

v^2 = u^2 + 2g ( x - xo)

or

x - xo = v0t + 1/2 gt^2

and what would my variables be

v0 = 8.82 m/s

g = 9.81 m/s2

We need to find 't' and add it to the 1.7seconds in the problem

That would be the total time, to hit the ground,

Plz. help!!!