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

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:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$   (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:

$0=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (3)  

Isolating $t$ from (3):  

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

$t=7.83s$ (5)  

Substituting (5) in (1):  

$x_{max}=30m/s.cos(40\°) (7.83)$   (6)  

Finally:

$x_{max}=180.102m$   (7)