Question

A rock is thrown off a cliff at an angle of 46° above the horizontal. The cliff is 120 m high. The initial speed of the rock is 21 m/s. (Assume the height of the thrower is negligible.)

(a)

How high above the edge of the cliff does the rock rise (in m)?

(b)

How far has it moved horizontally when it is at maximum altitude (in m)? m

(c)

How long after the release does it hit the ground (in s)?

(d)

What is the range of the rock (in m)?

(e)

What are the horizontal and vertical positions (in m) of the rock relative to the edge of the cliff at

t = 2.0 s,

t = 4.0 s,

and

t = 6.0 s?

(Assume the +x-direction is in the horizontal direction pointing away from the cliff, the +y-direction is up towards the sky, and x = y = 0 at the point from which the rock is thrown.)

mx(2.0 s)=

my(2.0 s)=

mx(4.0 s)=

my(4.0 s)=

mx(6.0 s)=

my(6.0 s)=

Answer 1

PART A:

At the highest point h, the vertical velocity becomes zero. i.e

So, the rock rises 11.643 m above the cliff.

PART B:

The time at maximum can be calculated as

The horizontal distance covered over this time is

So, the rock covers 22.486 m horizontally when it reaches its highest point.

PART C:

The final height can be taken as the height of the cliff with a negative sign ( taking the clifftop as the reference point).

We know

Putting values with proper unit and sign

After solving this quadratic equation we get

and

We were solving for a positive value.

So, the time of flight is 6.725 s.

PART D:

The range of the rock is therefore