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 m/s²)

Correct answer:

t =  4 s
x =  60 m

Step-by-step explanation:

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

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

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