Question

While delivering 125 kg blocks of ice to a local village, Kristoff and his family (and Sven too) come upon a cliff that is 5.7 m above them. To solve their problem, they build a catapult that will launch their blocks of ice with an initial velocity of 15 m/s and at an angle of 45 degrees above the ground.
Prove that the catapult will successfully launch the ice blocks up to the top of the cliff. Quantities you will need to solve for will be initial vertical velocity, horizontal velocity, the distance the catapult is from the cliff, and the amount of time it takes the ice to reach the cliff. Assume that they are experts and have arranged everything so that the ice blocks just barely reach the top of their parabolic path when they land at the top of the cliff and make a smooth landing. Draw a picture of the scene to help you visualize what is happening.


After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.

What is the rate of acceleration while the blocks slow to a stop?


How long do the blocks take to slow to a stop?


What is the amount of friction between the ice and the snowy ground?



How long do the blocks take to slow to a stop?


What is the amount of friction between the ice and the snowy ground?

Answer 1

Answer:

a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

b. -2.81 m/s²

c. 3.78 s

d. -351.25 N

Explanation:

a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.

For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.

The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².

For the block of ice, v = 15 m/s and Ф = 45°. So,

h = v²sin²Ф/2g

= (15 m/s)²sin²45/(2 × 9.8 m/s²)

= 225 (m/s)²(1/√2)²/19.6 m/s²

= 225 (m/s)²(1/2)/19.6 m/s²

= 112.5 (m/s)²/19.6 m/s²

= 5.74 m

Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

The graph is in the attachment.

b. What is the rate of acceleration while the blocks slow to a stop?

Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m

So, a =  (v² - u²)/2s

substituting the variables into the equation, we have

a =  ((0 m/s)² - (10.61 m/s)²)/2(20 m)

= - 112.57 (m/s)²)/40 m

= -2.81 m/s²

c. How long do the blocks take to slow to a stop?

Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop

So, making t subject of the formula,

t = (v - u)/a

substituting the values of the variables, we have

t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²

= -10.61 m/s/-2.81 m/s²

= 3.78 s

d.  What is the amount of friction between the ice and the snowy ground?

The frictional force, f = net force on block of ice

f = ma where  m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²

f = ma

= 125 kg(2.81 m/s²)

= -351.25 N