Gravitation

From the top of the 80m high tower, the body is thrown horizontally with an initial speed of 15 m/s. At what time and at what distance from the foot of the tower does the body hit the horizontal surface of the Earth? (use g = 10 ms-2)

Correct result:

t =  4 s
x =  60 m

Solution:

$$\begin{gathered} h=80\text{ m } \\ g=10{\text{m/s}}^2 \\ {v}_0=15\text{ m/s } \\ h={\displaystyle \frac{1}{2}}\cdot g{t}^2 \\ t=\sqrt{2\cdot h\mathrm{/}g}=\sqrt{2\cdot 80\mathrm{/}10}=4\text{ s} \end{gathered}$$

$x=v_0\cdot t=15\cdot 4=60\text{ m}$

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