When a ball is thrown with a velocity of 50   m   s - 1 at an angle 30 ° with the horizontal,…

The vertical component of velocity is, $50\sin 30°$therefore, $u=50\sin 30°$and when the ball will be thrown vertically the net displacement along the vertical direction of body will be zero, since body will again come to ground after time $t.$

Now using the equation of uniform motion, $s=ut+\frac{1}{2}a{t}^2$

$$\begin{gathered} \Rightarrow 0=\left(50\sin 30°\right)t+\frac{1}{2}\left(-g\right)t^2 \\ \Rightarrow 5t^2-\left(50\sin 30°\right)t=0 \cdots \cdots \left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right) \\ \Rightarrow t\left(5t-25\right)=0 \end{gathered}$$

Solving the equation for $t,$

$t=0, \text{and} t=5 \mathrm{s}$

Thus, the ball will remain in air for $5 \mathrm{s}$