Question

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

Answer 1

`y_m_a_x=10m`

Explanation:

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

`y_m_a_x= (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= (v^2*sin^2(\theta))/(2g)=((14^(2))*sin^2(90) )/(2*9.8)=(196*1)/(19.6)=10m`