# 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)

Result

t =  4 s
x =  60 m

#### Solution:

$h=80 \ \text{m} \ \\ g=10 \ \text{m/s}^2 \ \\ v_{0}=15 \ \text{m/s} \ \\ \ \\ h=\dfrac{ 1 }{ 2 } \cdot \ g t^2 \ \\ \ \\ t=\sqrt{ 2 \cdot \ h/g }=\sqrt{ 2 \cdot \ 80/10 }=4 \ \text{s}$
$x=v_{0} \cdot \ t=15 \cdot \ 4=60 \ \text{m}$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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