Paratrooper

After the parachute is opened, the paratrooper drops to the ground at a constant speed of 2 m/s, with the sidewinding at a steady speed of 1.5 m/s. Find:

a) the magnitude of its resulting velocity with respect to the ground,
b) the distance of his land from a tree over which it was dropped at an altitude of 800 m above the ground (AGL).

Result

v =  2.5 m/s
x =  600 m

Solution:

v1=2 m/s v2=1.5 m/s  v=v12+v22=22+1.52=52=2.5 m/sv_{1}=2 \ \text{m/s} \ \\ v_{2}=1.5 \ \text{m/s} \ \\ \ \\ v=\sqrt{ v_{1}^2 + v_{2}^2 }=\sqrt{ 2^2 + 1.5^2 }=\dfrac{ 5 }{ 2 }=2.5 \ \text{m/s}
h=800 m t=h/v1=800/2=400 s  x=t v2=400 1.5=600 mh=800 \ \text{m} \ \\ t=h/v_{1}=800/2=400 \ \text{s} \ \\ \ \\ x=t \cdot \ v_{2}=400 \cdot \ 1.5=600 \ \text{m}



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