Question

A man stands on top of a 25 m tall cliff. The man throws a rock upwards at a 25 degree angle above the horizontal with an initial velocity of 7.2 m/s. What will be the vertical position of the rock at t = 2 seconds? What will be the horizontal position of the rock at that time?

Answer 1

Step-by-step explanation:

It is given that,

Position of the man, h = 25 m

Initial velocity of the rock, u = 7.2 m/s

The rock is thrown upward at an angle of 25 degrees.

(a) Let y is the vertical position of the rock at t = 2 seconds. Using the equation of kinematics to find y as :

`y=-(1)/(2)gt^2+u\ sin\theta\ t+h`

`y=-(1)/(2)* 9.8(2)^2+7.2\ sin(25)* 2+25`

y = 11.48 meters

(b) Let x is the horizontal position of the rock at that time. It can be calculated as :

`x=u\ cos\theta * t`

`x=7.2\ cos(25)* 2`

x = 13.05 meters

Hence, this is the required solution.

Answer 2

a) vertical position at 2s: 50.68 m

b) horizontal position at 2s: 13.05 m

Step-by-step explanation:

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

x-component:

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

Where:

`V_(o)=7.2 m/s` is the rock's initial speed

`\theta=25\°` is the angle at which the rock was thrown

`t=2 s` is the time

y-component:

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

Where:

`y_(o)=25 m` is the initial height of the rock

`y` is the height of the rock at 2 s

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

Knowing this, let's begin with the answers:

a) Vertical position at 2 s:

`y=25 m+(7.2 m/s)sin(25\°) (2 s)-((9.8m/s^(2))(2)^(2))/(2)` (3)

`y=25 m+6.085 m-19. 6 m` (4)

`y=50.68 m` (5) This is the vertical position at 2 s

b) Horizontal position at 2 s:

`x=(7.2 m/s) cos(25\°) (2 s)` (5)

`x=13.05 m` (6) This is the horizontal position at 2 s