Kinematics

[d0minicz]

1.a) A stone is thrown vertically upwards from ground level at 21 m/s.
i) What is its height above the ground after two seconds? 22.4m
ii) What is the maximum height reached by the stone? 22.5m

b) If the stone is thrown vertically upwards from a cliff 17.5m high at 21 m/s:
i) how long will it take to strike the ground at the base of the cliff?
ii) what is the velocity of the stone when it hits the ground?

thanks.!

[kamil9876]

If we make positive upwards, then really we want the stone to be displaced by -17.5m.

  be careful with directions of acceleration and initialy velocity. Now solve for t.

partb:




sub t from part a.

[d0minicz]

A basketball is thrown vertically upwards with a velocity of 14 m/s. Find:
a) the time taken by the ball to reach its maximum height
b) the greatest height by the ball
c) the time taken for the ball to return to the point from which it is thrown
For c) , would the time just be doubled? As it has to cover twice the distance from the max height?

[Flaming_Arrow]

c) that is correct

[d0minicz]

thx man

An object is dropped from a point 100m above the ground. The acceleration due to gravity is . Find:
a) the time taken for the object to reach the ground
b) the velocity at which the object hits the ground

needing tips ; thx x]

[Flaming_Arrow]

a) d= 100 u=0 a=9.8 solve for t
b) solve for v

[d0minicz]

Hey im getting confused to which values should be negative and positive (u,v,a)
can anyone provide me with an explanation pleaseeeeeeeeeeeee

[Flaming_Arrow]

first assign which direction is positive

for en example

think of a ball throwing up, so we take up as positive

velocity will be positive, distance will be positice and acceleration will be negative because acceleration always acts downwards

[d0minicz]

ahhhh cheers mate

[d0minicz]

A particle travels in a straight line with a constant velocity of 4 m/s for 12 seconds. It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position. Find:
a) the acceleration of the particle
b) the time the particle is travelling back towards its original position

thanks...

[Flaming_Arrow]

d = 4* 12 = 48m

a)d = 48 t=20 u=-4 find a

[d0minicz]

can someone please show the working for it cant seem to get it
thanks !

[hyperblade01]

Well its constant acceleration so you use one of those formulas:

EDIT: I'm wrong, haven't done kinematics in a while :p

[Flaming_Arrow]

a) d = 48 t=20 u=-4 find a


48 = -80 + 0.5 * a *400

a = -16/25

[hyperblade01]

That's what I got, but its -.64m/s^2


EDIT:

Have to use:







EDIT 2: Even though its right, I think I used the wrong equation...