Up and down motion

We throw the body from a height h = 5 m above the Earth vertically upwards v0 = 10 m/s. How long before we have to let the second body fall freely from the same height to hit the Earth at the same time?

Correct result:

t =  1.4445 s

Solution:

h=5 m v0=10 m/s g=9.81 m/s2  t1=v0/g=10/9.81=10009811.0194 s  h1=h+12 g t12=5+12 9.81 1.01942=990598110.0968 m  h1=12 g t22  t2=2 h1/g=2 10.0968/9.811.4347 s  t3=2 h/g=2 5/9.811.0096 s  t+t3=t2+t1  t=t2+t1t3=1.4347+1.01941.0096=1.4445 s



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