Question

The maximum height of a projectile thrown at 5 m/s at an angle of 300 with horizontal is​

Answer 1

Answer:

The maximum height of a projectile thrown at 5 m/s at an angle of $30^{0}$ with horizontal is​    0.3125 m.  

Explanation:

For a projectile motion, the equation of maximum height is given by,

                            $h_{max} = \frac{u^{2} \sin ^{2} \theta}{2 g}$    ….. ( 1 )

Where,

             u - Initial velocity

             θ - angle of projection

             g - Acceleration due to gravity

Given,

            u = 5 m/s

            θ = $30^{0}$            ;   {   $sin 30^{0} = \frac{1}{2}$    }

            $g = 10 m/s^{2}$

Substituting above values in to equation ( 1 ) ,

 equation (1) becomes,      

                           $h_{max} = \frac{(5)^{2} \times\left(\frac{1}{2} \right)^{2}}{2 \times 10}$

                                     $= \frac{25\times \frac{1}{4} }{20}$    

                                     $\begin{array}{l}=0.3125 \mathrm{~m}\end{array}$