# A student throws a water balloon vertically downward from the top of a building? [see below]

## A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 8.00 m/s. Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand. a) What is its speed after falling for 2.00 s? b) How far does it fall in 2.00 s? c) What is the magnitude of its velocity after falling 10.0 m? How do I sketch a-1, V-t, and yet graphs for the motion of the balloon?

Jul 4, 2018

#### Answer:

a. $v = 27.6 \frac{m}{s} \text{ }$b. $d = 35.6 m \text{ }$ c. $v = 16.1 \frac{m}{s}$

#### Explanation:

a. Use the motion formula $v = u + a \cdot t$.
$v = 8.00 \frac{m}{s} + 9.81 \frac{m}{s} ^ \cancel{2} \cdot 2.00 \cancel{s} = 27.6 \frac{m}{s}$

b. Use the motion formula $d = u \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$.
$d = 8.00 \frac{m}{\cancel{s}} \cdot 2.00 \cancel{s} + \frac{1}{2} \cdot 9.81 \frac{m}{s} ^ 2 \cdot {\left(2.00 s\right)}^{2}$

$d = 16.0 m + \frac{1}{2} \cdot 9.81 \frac{m}{\cancel{{s}^{2}}} \cdot 4.00 \cancel{{s}^{2}}$

$d = 16.0 m + 19.62 m = 35.6 m$

c. Use the motion formula ${v}^{2} = {u}^{2} + 2 \cdot a \cdot d$.

${v}^{2} = {\left(8.00 \frac{m}{s}\right)}^{2} + 2 \cdot 9.81 \frac{m}{s} ^ 2 \cdot 10.0 m$

${v}^{2} = 64.0 {m}^{2} / {s}^{2} + 196.2 {m}^{2} / {s}^{2} = 260.2 {m}^{2} / {s}^{2}$

$v = \sqrt{260.2 {m}^{2} / {s}^{2}} = 16.1 \frac{m}{s}$