Biomechanics
posted by Jack .
A ball is thrown vertically upwards in the air at a velocity of 50(m/s) and reaches a maximum height of 20(m). Calculate the time when the ball reaches the 10(m) mark, while going up and coming down. Use the following equation of motion:
y(f) = y(i) + v(i)t+1/2at^2
y(f) and y(i) = final and initial position, v(i) = initial velocity, a = acceleration, and t = time, respectively.

velocity is 509.8t.
v=0 at top of trajectory, at t=50/9.8=5.10 sec.
y = yi + 50t  4.9t^2, so at t=5.10,
20 = yi + 50(5.10)  4.9(5.10)^2
yi = 107.55
eh? thrown from below ground?
Yet it makes sense, since it was thrown upward with a velocity of 50 m.s, yet only made it 20m high. No mention was made of a horrific air resistance.
Anyway, assuming the above,
y = 107.55 + 50t  4.9t^2
solve for t when y=10:
t = 3.67 or 6.53
If I botched it, make the fix.