Physic 20 Help

[AznDragon2004]

I need help with a question on physics.


A rock is thrown horizontally at a speed of 5.00 m/s, from the top of a 10.0m high cliff.

Calculate:

a) The time taken to fall

b) The horizontal distance traveled

c) The final velocity of the rock

d) At what angle does it hit the ground

Thanks

[Impreza]

DRAW A DIAGRAM!!!!!! it might help you visualize things a bit better

[djayz]

dont you guys have those work booklets ?...super helpful

if i find mine soon ill help you out

[Shaolin]

Diagrams and a formula sheet.. that's all you need to pass Physics 20.

[Zero102]

dD = V0dT + 1/2AdT^2
We'll consider this in the vertical direction only, to solve part a.

V0 = 0, since all of the velocity was horizontal.
now, we are left with dD = 1/2AdT^2
We know that A = 9.81m/s, and dD is 10.0m (the height of the cliff)
So, 2dD/A = dT^2
or sqrt(2dD/A) = T
Or, T = 1.428s
So, horizontal distance travelled would be 5 * 1.428 or 7.14m
The final velocity of the rock would be the result of vector addition of the 2 components of it's velocity,
so, |Vf| = sqrt(Vh^2 + Vv^2),
or |Vf| = sqrt(25 + 14.008^2) or 14.874m/s

The angle it hits the ground at takes a little bit of trig, but the horizontal velocity is 5m/S, and the vertical velocity is 14.008m/s, so it will arrive at 19.643 degrees from the vertical.

I don't have the time to check my work, so maybe it's right, or maybe I've been awake for too long, but this should give you something to check your work against.
There really isn't enough here to get you full marks, so it is only really useful for checking work.

[AznDragon2004]

Originally posted by Zero102
dD = V0dT + 1/2AdT^2
We'll consider this in the vertical direction only, to solve part a.

V0 = 0, since all of the velocity was horizontal.
now, we are left with dD = 1/2AdT^2
We know that A = 9.81m/s, and dD is 10.0m (the height of the cliff)
So, 2dD/A = dT^2
or sqrt(2dD/A) = T
Or, T = 1.428s
So, horizontal distance travelled would be 5 * 1.428 or 7.14m
The final velocity of the rock would be the result of vector addition of the 2 components of it's velocity,
so, |Vf| = sqrt(Vh^2 + Vv^2),
or |Vf| = sqrt(25 + 14.008^2) or 14.874m/s

The angle it hits the ground at takes a little bit of trig, but the horizontal velocity is 5m/S, and the vertical velocity is 14.008m/s, so it will arrive at 19.643 degrees from the vertical.

I don't have the time to check my work, so maybe it's right, or maybe I've been awake for too long, but this should give you something to check your work against.
There really isn't enough here to get you full marks, so it is only really useful for checking work.


Thanks

[AznDragon2004]

Originally posted by Zero102
dD = V0dT + 1/2AdT^2
We'll consider this in the vertical direction only, to solve part a.

V0 = 0, since all of the velocity was horizontal.
now, we are left with dD = 1/2AdT^2
We know that A = 9.81m/s, and dD is 10.0m (the height of the cliff)
So, 2dD/A = dT^2
or sqrt(2dD/A) = T
Or, T = 1.428s
So, horizontal distance travelled would be 5 * 1.428 or 7.14m
The final velocity of the rock would be the result of vector addition of the 2 components of it's velocity,
so, |Vf| = sqrt(Vh^2 + Vv^2),
or |Vf| = sqrt(25 + 14.008^2) or 14.874m/s

The angle it hits the ground at takes a little bit of trig, but the horizontal velocity is 5m/S, and the vertical velocity is 14.008m/s, so it will arrive at 19.643 degrees from the vertical.

I don't have the time to check my work, so maybe it's right, or maybe I've been awake for too long, but this should give you something to check your work against.
There really isn't enough here to get you full marks, so it is only really useful for checking work.

For the first question you used the formula d=vit + at^2/2. What happen to your t infront of the vi?

[Audio_Rookie]

t is multiplied by v, and since v is 0 anything times that would = 0. so you can just cross out vi with t.

U can find vertical velocity first by going.
vf^2 = vi^2 + 2ad
then since vi = 0m/s you can just cross out vi^2 and get
vf^2 = 2ad where a = 9.81m/s^2 and d = 10.0m
vf^2 = 2(9.81m/s^2)(10.0m)
vf^2 = 196.2m^2/s^2 then square root each side and get

vf = 14.007m/s ,

time can be found with d = 1/2(vf + vi)t
then you get 2d = (vf+vi)t , and then dividing you get
t = 2d/(vf+vi)
where d = 10.0m , vf = 14.007m/s, and vi = 0m/s
then you get t = 1.428s, or with significant digits 1.43s

it travels 5.0m/s horizontally, and thankfully horizontal velocity dosn't change....well atleast not in this physics.
use v = d/t and then get d = vt, d = (5.0m/s) x (1.428s). d = 7.14m


we know vertical velocity = 14.007m/s and horizontal velocity = 5.0m/s. now we can use the good old pythagerous where a^2 + b^2 = c^2 to find final velocity

= (14.007m/s)^2 + (5.0m/s)^2 = c^2
= 221.196m^2/s^2 = c^2 now square root both sides.
= c = 14.87266m/s........ since you probobly are using significant digits you only use 3 digits in answer and round where neccessary = 14.9m/s


the angel of the rock we can use tan0 = opp/adj
then get tan0 =(14.007m/s) / (5.0m/s)
tan0 = 2.8014. 2nd function of tan(2.8014) = 70.3552 degrees, wich when rounding for significant digits gives you 70.4 degrees (from the horizontal)

hope this helps some more.
hope i remembered formulas correctly, but i got roughly same answer as Xero, so they should be correct.