Question

You throw a water balloon straight up with a velocity of 14 m/s. What is it’s max height?

Answer 1

Answer:

$y_m_a_x=10m$

Step-by-step explanation:

The max height of an object can be calculated using one the basic projectile motion equations:

$y_m_a_x= \frac{v^2*sin^2(\theta)}{2g}$

Where:

$g=Gravitational\hspace{3}constant\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}object\\\theta=Launch\hspace{3}angle$

In this case, let's asumme g=9.8m/s^2, besides θ=90, because the water ballon was threw straight up, so:

$y_m_a_x= \frac{v^2*sin^2(\theta)}{2g}=\frac{(14^{2})*sin^2(90) }{2*9.8}=\frac{196*1}{19.6}=10m$