Q. A man on the edge of a cliff H = 40 m high throws a ball directly upward. It returns past him 1.8 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)(adsbygoogle = window.adsbygoogle || []).push({});

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!!!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Ball thrown from cliff

**Physics Forums | Science Articles, Homework Help, Discussion**