an arrow is launched straight up from the ground with an initial velocity of 23.4 m/s. how long until it reaches its highest point?

Explanation:

Given that,

The initial velocity of an arrow is 23.4 m/s

We need to find the time taken by it to reach its maximum height. Let the maximum height is h.

Using third equation of motion to find height first.

At maximum height, v = 0

a = -g

Let t is the time to reach its maximum height. So, it can be calculated using second equation motion as follows :

Hence, it will take 2.35 seconds to reach its maximum height.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it