Avery strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 12.0 m/s. how long does he have to get out of the way if the shot was released at a height of 2.10 m, and he is 1.90 m tall?

The shot has a constant downward acceleration of . Since acceleration is constant, the velocity of the ball is given by

where is the shot's vertical velocity at time and is its initial vertical velocity. We're given that . So

gives the shot's velocity at time .

Because acceleration is constant, we also know that

where is the average vertical velocity, is the vertical displacement at time , and is the initial displacement at . We're given an initial displacement of , so the displacement of the shot is

The shotputter is 1.90 m tall, a difference of 0.20 m from the initial height of the throw. So we want to find the time for the ball to reach a displacment of -0.20 m:

root word, prefixes, and suffixes

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.