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 m g=10 m/s2 v0=15 m/s  h=12 gt2  t=2 h/g=2 80/10=4 sh=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=v0 t=15 4=60 mx=v_{0} \cdot \ t=15 \cdot \ 4=60 \ \text{m}



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