Question

" A 4 kg ball is thrown at an angle of 40 degrees above the horizon from the top of a 150 m cliff. How far from the base of the cliff does it land of the initial velocity is 30 m/s

Answer 1

Answer: 180.102m

Step-by-step explanation:

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

x-component:

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

Where:

`V_(o)=30m/s` is the ball's initial speed

`\theta=40\°` is the angle above the horizon

`t` is the time since the ball is thrown until it hits the ground

y-component:

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

Where:

`y_(o)=150m` is the initial height of the ball

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

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

Knowing this, let's begin with the calculations:

Firstly, we have to find the time the ball elapsed traveling. So, we will use equation (2) with the conditions given above:

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

Isolating
`t` from (3):

`t=(V_(o))/(g)(sin\theta+\sqrt{{sin\theta}^(2)+2(g.y_(o))/(V_(o)^(2))})` (4)

`t=7.83s` (5)

Substituting (5) in (1):

`x_(max)=30m/s.cos(40\°) (7.83)` (6)

Finally:

`x_(max)=180.102m` (7)