Question

You throw a glob of putty straight up toward the ceiling, which is 3.00m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.20m/s .

What is the speed of the putty just before it strikes the ceiling?Express your answer with the appropriate units.



How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.

Answer 1

Assume that the air resistance on this glob of putty is negligible. Let
`g = \rm -9.81\; m \cdot s^(-2)`.

• The velocity of the glob of putty is approximately
  `5.08\; \rm m \cdot s^(-1)` right before it hits the ceiling.
• It would take about
  `0.420\; \rm s` for the glob of putty to reach the ceiling.

Step-by-step explanation:

Assume that the air resistance on this glob of putty is negligible. Because of gravity, the glob of putty would accelerate downwards at a constant
`g = \rm -9.81\; m \cdot s^(-2)`. Since the acceleration points downwards, its value should be negative.

Consider the equation for an object under constant acceleration, where the time taken is unknown.

`\displaystyle v^2 - u^2 = 2\, a \cdot x`,

where

• 
  `v` is the final velocity of the object. In this case, the value of
• 
  `u` is the initial velocity of the object. In this case,
  `u = 9.20\; \rm m \cdot s^(-1)`. Note that in this case,
• 
  `a` is the acceleration on the object. In this case,
  `a = g = -9.81\; \rm m \cdot s^(-2)`.
• 
  `x` is the displacement of the object (during the time when its velocity changed from
  `u` to
  `v`.) In this case,
  `x = 3.00\; \rm m`. The value of

Rearrange the equation for
`v`:

`v^2 = 2\, a \cdot x + u^2`.

`v = √(2\, a \cdot x + u^2)`.

Calculate the value of
`v`:

`\begin{aligned}v &= √(2\, a\cdot x + u^2) \\ &= \sqrt{2 * \underbrace{(-9.81)}_(g) * \underbrace{3.00}_(x) + {\underbrace{9.20}_(u)}^2} \\ &\approx 5.08\; \rm m \cdot s^(-1) \end{aligned}`.

It is assumed that there's no air resistance on the glob of putty. As a result, the acceleration of the putty would be constant. The duration of this motion can be found with the equation:

`\displaystyle t = (\Delta v)/(a) = (v - u)/(a)`,

where again,

• 
  `v` is the final velocity of the blob of putty,
• 
  `u` is the initial velocity of the blob of putty, and
• 
  `a` is the acceleration of the blob of putty.

Based on the previous conclusion,
`v \approx 5.08\; \rm m \cdot s^(-1)`. Once again,
`u = 9.20\; \rm m \cdot s^(-1)` and
`a = g = -9.81\; \rm m \cdot s^(-2)`. (Keep in mind that the value of the acceleration of the blob is negative.)

`\begin{aligned} t &= (v - u)/(a) \\ &= (5.08 - 9.20)/(-9.81) \\ &\approx 0.420 \; \rm s\end{aligned}`.

Note that the inputs
`u`,
`g`, and
`x` here are all in standard units. As a result, the value of
`v` and
`t` from the formula would also be in standard units.