# A stone is thrown from the top of a building upward at an angle of 30.0∘ to the horizontal and with an initial speed of 20.0 m+s. If the height of the building is 45.0 m, how much time is it in seconds before the stone hits the ground g =9.8 m + sec2

Dear Student,

Please find below the solution to the asked query:

Given information is,

Initial Speed is *u=20 m/s*, angle of projection is *θ = 30 ^{0}*, height of the building

*h = 45 m*and acceleration due to gravity is

*g = 9.8 m/s*.

^{2}Here the final position is below than the initial position. So, the displacement is negative. Acceleration is in opposite direction to the initial velocity, so that is also negative. Let the time taken by the body to reach the ground is

*t*.

From the kinematics equation,

${S}_{y}={U}_{y}t+\frac{1}{2}{a}_{y}{t}^{2}\Rightarrow -45=20t-\frac{1}{2}\times 9.8\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4.9{t}^{2}-20t-45=0\phantom{\rule{0ex}{0ex}}t=\frac{-\left(-20\right)\pm \sqrt{{\left(-20\right)}^{2}-4\left(4.9\right)\left(-45\right)}}{2\times 4.9}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{20\pm \sqrt{400+882}}{9.8}=\frac{20\pm \sqrt{1282}}{9.8}=\frac{20\pm 35.8}{9.8}\phantom{\rule{0ex}{0ex}}timecannotbenegative.So,\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{20+35.8}{9.8}=\frac{55.8}{9.8}\phantom{\rule{0ex}{0ex}}\Rightarrow t=5.7s\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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