Chapter 1: Q38E (page 61)

You throw a glob of putty straight up toward the ceiling,which is 3.60 m above the point where the putty leaves your hand.The initial speed of the putty as it leaves your hand is 9.50 m/s.
(a) What is the speed of the putty just before it strikes the ceiling?
(b) How much time from when it leaves your hand does it take theputty to reach the ceiling?

Expert verified

(a) The speed of the putty just before it strikes the ceiling is 4.44m/s., and

(b) The time putty takes to reach the ceiling is t =0.52s

Step by step solution

The initial speed of putty is $u = 9.50 m / s$

The distance between your hand and the ceiling is $s = 3.60 m$

The acceleration due to gravity is acting opposite direction, therefore, $a = - 9.8 m / s$

The final speed (v) can be determined using Newton’s third law of motion as,

$v^{2} = u^{2} + 2 a s$

Substituting the values in the above expression,

$v^{2} = {(9.50 m / s)}^{2} - 2 \times (9.8 m / s^{2}) \times (3.60 m) v^{2} = 19.69 m^{2} / s^{2} v = 4.44 m / s$

thus, the speed of the putty just before it strikes the ceiling is 4.44m/s.

If we assume that t is the required time, then applying Newton’s first law of motion, we can write,

$v = u + a t$

Substitute the values in the above expression,

$\begin{array}{rcl} t & = & \frac{4.44 m / s - 9.50 m / s}{- 9.80 m / s^{2}} \\ = & 0.52 s \end{array}$

Thus, the time putty takes to reach the ceiling is $t = 0.52 s$ .

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