Question

A rock is propelled off a pedestal that is 10 meters off the level ground. The rock leaves the pedestal with a speed of 18 meters per second at an angle above the horizontal of 20 degrees. How high does the rock get, and how far downrange from the pedestal does the rock land?

Answer 1

Answers:

a) How high does the rock get?=1.933m

b)How far downrange from the pedestal does the rock land?=21.25m

Explanation:

This situation is a good example of projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being their main equations as follows:

x-component:

`x=V_(o)cos\theta t` (1)

`V_(x)=V_(o)cos\theta` (2)

Where:

`V_(o)=18m/s` is the rock's initial speed

`\theta=20\°` is the angle

`t` is the time since the rock is propelled until it hits the ground

y-component:

`y=y_(o)+V_(o)sin\theta t-(gt^(2))/(2)` (3)

`V_(y)=V_(o)sin\theta-gt` (4)

Where:

`y_(o)=10m` is the initial height of the rock

`y=0` is the final height of the rock (when it finally hits the ground)

`g=9.8m/s^(2)` is the acceleration due gravity

Knowing this, let's begin with the anwers:

a) How high does the rock get?

Here we are talking about the maximun height
`y_(max)` the rock has in its parabolic motion. This is fulfilled when
`V_(y)=0`.

Rewritting (4) with this condition:

`0=V_(o)sin\theta-gt` (5)

Isolating
`t`:

`t=(V_(o)sin\theta)/(g)` (6)

Substituting (6) in (3):

`y_(max)=y_(o)+V_(o)sin\theta((V_(o)sin\theta)/(g))-(1)/(2)g((V_(o)sin\theta)/(g))^(2)` (7)

`y_(max)=(V_(o)^(2)sin^(2)\theta)/(2g)` (8)

Solving:

`y_(max)=((18m/s)^(2)sin^(2)(20\°))/(2(9.8m/s^(2)))` (9)

Then:

`y_(max)=1.933m` (10) This is the maximum height the rock has.

b) How far downrange from the pedestal does the rock land?

Here we are talking about the maximun horizontal distance
`x_(max)` the rock has in its parabolic motion (this is fulfilled when
`y=0`):

`0=y_(o)+V_(o)sin\theta t-(gt^(2))/(2)` (11)

Isolating
`t` from (11):

`t=(2V_(o)sin\theta)/(g)` (12)

Substituting (12) in (1):

`x_(max)=V_(o)cos\theta ((2V_(o)sin\theta)/(g))` (13)

`x_(max)=(V_(o)^(2)(2cos\theta sin\theta))/(g)` (14)

Knowing
`sin(2\theta)=2cos\theta sin\theta`:

`x_(max)=(V_(o)^(2)sin2\theta)/(g)` (15)

Solving:

`x_(max)=((18m/s)^(2)sin2(20))/(9.8m/s^(2))` (16)

Finally:

`x_(max)=21.25m` (17)