Question

A stone is thrown from a point A at the top of a building of height H = 14 m with an initial downward velocity v - -2m/s. What is the time needed for this stone to reach a point B at a height h = 4 m above the ground? Neglect the effect of air resistance, and use g = 10 m/s

Answer 1

Given

Total height of the building, H=14 m

The initial velocity is v=-2 m/s

The point B is at height, h=4 m.

To find

The time need by the stone to reach point B.

Step-by-step explanation

Using the equation of kinematics,

`y=ut+(1)/(2)gt^2`

where y is the distance covered.

t is the time taken

Thus,

`\begin{gathered} H-h=vt+(1)/(2)gt^2 \\ \Rightarrow14-4=2t+(1)/(2)*10* t^2 \\ \Rightarrow10=5t^2+2t \\ \Rightarrow5t^2+2t-10=0 \\ \Rightarrow t=(-2\pm√(4+200))/(10) \\ \Rightarrow t=1.28\text{ s \lparen considering only the positive term\rparen} \end{gathered}`

Conclusion

The time is 1.28 s