Question

A water balloon is thrown horizontally with a speed of 8.31 m/s from the roof of a building of height H=23.0 m. How far does the balloon travel horizontally before striking the ground?

Answer 1

Given:

A water balloon is thrown horizontally with a speed of 8.31 m/s from the roof of a building of height H=23.0 m.

To find:

Distance travelled horizantally before striking the ground ?

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(0,5)(5,4.5)(5,0)\put(0,0){\line(0,1){5}}\put(-1,0){\line(1,0){7}}\put(0,5){\vector(1,0){3.5}}\multiput(-0.5,0)(0.5,0){13}{\qbezier(0,0)(0,0)(-0.3,-0.3)}\put(-1,2.5){\bf{h}}\put(4,4.9){ \bf{u = u_x} }\put(2.3,-1){\bf{x\:=\:?}}\end{picture}$

Calculation:

This is an example of Height - to - Ground PROJECTILE:

Since, the object is thrown horizantally, the Y axis component of velocity of object will be zero.

Let time taken to reach ground be "t":

$\therefore \: h = u_{y}t + \dfrac{1}{2}g {t}^{2}$

$\implies \: 23 = (0)t + \dfrac{1}{2}(10){t}^{2}$

$\implies \: 23 = 0 + \dfrac{1}{2}(10){t}^{2}$

$\implies \: 23 = 5{t}^{2}$

$\implies \:{t}^{2} = \dfrac{23}{5}$

$\implies \:{t}^{2} = 4.6$

$\implies \:t = \sqrt{4.6}$

$\implies \:t = 2.14 \: sec$

Now, let the distance travelled horizantally be "x"

$\therefore \: x = u_{x} \times t$

$\implies \: x = 8.31 \times 2.14$

$\implies \: x = 17.78 \: m$

So, distance travelled horizantally before striking ground is 17.78 m.

Answer 2

Answer:

Explanation: idk