# Question 08187

Jul 6, 2015

Time needed for the ball to fall: 1 s.
Horizontal range: 0.8 m.

#### Explanation:

FULL QUESTION

A little ball hanged by a string is made to oscillate. After that, when it passes the the lowest point it can pass in its motion the string is cut.

If the ball moved in a height of 5m from the ground in a velocity of ${\text{0.8 ms}}^{- 1}$ at this moment,

• What is the time that takes for the ball to fall on the ground?
• What is the horizontal range passed by the ball?

The idea behind this problem is that, at the moment the string is cut, the ball will begin to act as a projectile launched horizontally with an initial velocity equal to that had at that point.

In your case, when the string is cut, the ball will start with an initial velocity of 0.8 m/s. Since it is launched horizontally, the vertical component of this velocity will be equal to zero.

That happens because the ball is launched at an${0}^{\circ}$ angle with the horizontal.

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cdot \cos \left({0}^{\circ}\right) = {v}_{0} \\ {v}_{0 y} = {v}_{0} \cdot \sin \left({0}^{\circ}\right) = 0\end{matrix}\right.$

Vertically, the ball will move under the influence of gravity. Knowing what its vertical displacement is, you can write

$- h = {\underbrace{{v}_{0 y}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$, where

$t$ - the time of flight.

This means that $t$ is equal to

t = sqrt( (2 * h)/g) = sqrt( (2 * 5cancel("m"))/(10cancel("m")/"s"^(2))) = color(green)("1 s")

The ball touches the ground 1 second after the string is cut.

Horizontally, the movement of the ball is not affected by any force, which means that you can write

$d = {v}_{0 x} \cdot t$, where

$t$ - the time of flight.

This means that the range of the ball is

d = underbrace(v_(0x))_(color(blue)(=v_0)) * t = v_0 * t = 0.8"m"/cancel("s") * 1cancel("s") = color(green)("0.8 m")#

The ball will drop 0.8 m away from the point in which it was when the string was cut.