Use the following information to answer questions 4 and 5:
A rock is launched vertically into the air at a velocity of 14.75 m/s.
4. Toby claims that the rock must come to rest before it can fall back towards the ground. Is Toby
correct?
A. Toby is correct because the rock is experiencing a negative acceleration, causing its negative
velocity to increase until the rock reaches a velocity of O m/s before becoming positive.
B. Toby is correct because the rock is experiencing a negative acceleration, causing its positive
velocity to decrease until the rock reaches a velocity of O m/s before becoming negative.
C. Toby is incorrect because the rock is experiencing a positive acceleration, causing its positive
velocity to increase in magnitude.
D. Toby is incorrect because the rock is experiencing a negative acceleration, causing its positive
velocity to increase in magnitude.
5. Calculate the time it takes for the rock to reach its maximum height.
A. 1.50 seconds
B. 2.47 seconds
C. 3.00 seconds
D. 4.94 seconds

B. Toby is correct because the rock is experiencing a negative acceleration, causing its positive velocity to decrease until the rock reaches a velocity of O m/s before becoming negative.

Question 5

At the maximum height, velocity is 0, so:

v = v₀ - gt

0 = 14.75 - 9.8t

t = 14.75/9.8

t = 1.5 s (OPTION A)

Answer Link

JackelineCasarez JackelineCasarez · Answer 2 · 2021-12-18T18:18:48+01:00

4). In the given situation, the assertion that could be made regarding Toby would be:

B). Toby is correct because the rock is experiencing a negative acceleration, causing its positive velocity to decrease until the rock reaches a velocity of O m/s before becoming negative.

5). The time that would be taken by the rock to attain its greatest height would be:

1.5 second

4). Tobby correctly states as rock would be undergoing a -ve acceleration which leads its +ve velocity to fall by the time rock attains the velocity of 0 m/s.

Thus, option B is the correct answer.

5). Given that,

Initial velocity([tex]v_{0}[/tex]) = 14.75 m/s

As we know,

Velocity remains 0 m/s at the greatest height,

So,

Velocity(v) [tex]= v_{0} - gt[/tex]

where

[tex]0 = 14.75 - 9.8t[/tex]

⇒ [tex]t = 14.75/9.8[/tex]

∵ [tex]t = 1.5 s[/tex]

Learn more about "Velocity" here:

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