I need the steps too please 1- A ball is thrown vertically upward with a speed o

Question

I need the steps too please


1- A ball is thrown vertically upward with a speed of
25.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the
ball take to hit the ground after it reaches its highest
point? (d) What is its velocity when it returns to the
level from which it started?


2-A small mailbag is released from a helicopter that is
descending steadily at 1.50 m/s. After 2.00 s, (a) what
is the speed of the mailbag, and (b) how far is it below
the helicopter? (c) What are your answers to parts (a)
and (b) if the helicopter is rising steadily at 1.50 m/s?


3-A baseball is hit so that it travels straight upward
after being struck by the bat. A fan observes that it
takes 3.00 s for the ball to reach its maximum height.
Find (a) the ball

Explanation / Answer

At its maximum height (ie just before it starts to fall back down) its velocity is zero. So for the rise its final velocity (v) = 0 m/s and its initial velocity (u) = 25.0 m/s The only acceleration is that due to gravity but direction must be considered. Take the upwards direction as positive so the acceleration = -g ie -9.8m/s^2 The relevant formula to use is v^2 = u^2 + 2as and you can use this to find the height to which the ball rises. The ball starts at a height of 2.0 m so that will probably have to be added on or alternatively you could say that it rises so many meters from where it is thrown. For part (b) just use v = u + at and solve for t. v, u and a are the same as for part (a) (c) This time the ball is hitting the ground which is 2.0m below where it started, so its displacement is -2.0m (since up is being taken as positive. The relevant formula is s = ut + 1/2 at^2. At its greatest height the balls velocity is zero , so for the fall to the ground, its initial velocity (u) = 0. This then simplifies the equation you have to solve. Reject the negative value of t because the time cannot be negative. (d) When the ball returns to the level from which it started its displacement is zero because it hasn't changed. The time for the fall to this level is the same as the time it took to rise to its highest point. So v = u + at is the relevant formula, once again just considering the fall so the initial velocity = 0m/s. You'll notice something about the magnitude of this velocity, something to remember for future questions like this, but be careful with direction because all vector quantities always have a magnitude and a direction.

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