projectile motion (1 Viewer)

[Danger]

1. A stone is thrown upwards from the edge of a cliff 25m high at 18m/s. On the way down it misses the cliff-top and falls to the ground below. Calculate:
(a) its total time in the air
(b) its speed just before it hits the ground.

Answers:
(a) 4.75s
(b) 28.5 m/s

Any help would be greatly appreciated. Also, anyone have any tips for doing projectile motion questions?

 

[watatank]

calculate time it is up in the air before it goes down first.

v = u + at,

0 = 18 -9.8*t

solve for t

then you will have to figure out the height that it reaches before it starts to fall

s = ut + 0.5at^2

you've got all the variables needed so solve for s. note that this 's' is the height above the cliff

then when it falls

v^2 = u^2 + 2as

v^2 = 0+ 2*9.8*(height from top)

your height in this case will be the height above which you threw it + the 25m above the cliff.

edit: i did the second question first. to answer the first question you will have to put that velocity you found into v = u + at and solve for t. add that to the first t you found and you've got the answer.

sorry the solution is all over the place. hope that helps.

tips: write down all the equations, write down everything you know and solve for the unknowns...define your positive direction like acceleration due to gravity obviously goes towards the earth, but think about which way a projectile is moving for example.

 

[crono26]

just like what wata tank said

initally
v=u+at
0=18+9.8t
t=1.83
therefore height above cliff
h=1/2at^2
=16.53m
therefore total height = 16.53+25
=41.5
total falling time
h=1/2at^2
41.5=1/2x9.8xt^2
t=2.91 (to fall)

therefore total time =up + down
=1.83+2.911
=4.74 (rounding is why i didnt get 4.75)

final speed when hitting the ground

v^2=u^2+2as
v^2=0+(2x9,8x41.5) (as it only begins to gain speed from its maximal height you dont need to worry about the inital vertical gain when he throws it)
v^2=813.4
v=28.5m/s

there you have it

 

[Danger]

thanks to both of you