91影视

The given information is,

$$\begin{gathered} \mathrm{Acceleration}\left(\mathrm{a}\right)=\mathrm{Gravitational}\mathrm{Acceleration}=9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Displacement}\left(\mathrm{y}\right)=50\mathrm{m} \\ \mathrm{Initial}\mathrm{Velocity}\left({\mathrm{v}}_0\right)=0\mathrm{m}/\mathrm{s} \end{gathered}$$

The problem deals with the concept of free fall. In free fall the object experiences the motion where gravity is the only force acting upon it. This concept can be used to find the velocity at which the ball must be thrown to reach a certain height. Once thrown, it would be acted upon by gravitational acceleration only. Hence, at maximum height,, the final velocity would be zero. Using these conditions, the initial velocity and the time for which ball would be in the air can be found.

Formula:

$$\begin{gathered} v^2=v_0^2+2ay \\ v=v_0+at \end{gathered}$$

Using the third equation for kinematic,

$$\begin{gathered} v^2={v_0}^2+2ay \\ {v}^2=0+2脳9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}脳50\mathrm{m} \\ {v}^2=\sqrt{980}{\mathrm{m}}^2/{\mathrm{s}}^2 \\ v=31.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Initial speed for the ball to travel 50mvertically is 31.3 m/s

The total time taken by the ball to reach the maximum height can be found as below

$v=v_0+at$

The final velocity at the maximum height would be zero. 鈥渁鈥� is the acceleration, and in this case, it is taken as 鈥�- g鈥�. It is negative because the upward direction is taken as positive.

$$\begin{gathered} 0=v_0-gt \\ t=v/g \end{gathered}$$

But the time duration of the ball in the air, so if it takes 鈥榯鈥� seconds to reach at maximum height, it will take 鈥榯鈥� seconds to reach to ground from the maximum height.

So

$$\begin{gathered} T=t+t \\ =2t \\ T=2脳\frac{v}{g} \\ T=2脳\frac{31.3\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ T=6.38 \\ 鈮�6.4\sec \end{gathered}$$