Projectile Motion Question (1 Viewer)

[velox]

my friend gave me this question but i keep getting the wrong answer

A stone is thrown vertically upwards with a velocity of 29.4 m/s from the edge of a cliff 78.4 m high. The stone falls so that it just misses the edge of the cliff and falls to the ground at the foot of the cliff. Determine the time taken by the stone to reach the ground.

Any ideas?

 

[Uncouth]

I posted a message saying that I got the wrong answer and I then figured out I left something out, silly me.

First of all, work out time to get to the top, v = 0.

v = u + at
0 = 29.4 + 9.8 * t
t = 3 s

Now find the distance between the top of the throw and the bottom of the cliff, and to do that you need to find the distance between the edge of the cliff and the top of the throw.

s = ut + ½at²
s = 0 + ½ * 9.8 * 3²
s = 44.1 m

Add that to the cliff height to find total height.

44.1 + 78.4 = 122.5 m

Find the time from top of the throw to the bottom of the cliff.

s = ut + ½at²
122.5 = 0 + ½ * 9.8 * t²
t = 5 s

Add the times together for the total time.

3 + 5 = 8 s.

So that's the answer. I hope. There are probably better ways to do it, but that's what I do in university physics.

 

[velox]

arggh i dont understand it, maybe i should read it again later tonight.
In the first section of working, why is the acceleration 9.8ms? Shouldnt it be -9.8ms?

 

[withoutaface]

Yes it should be. In maths however you take it as 9.8 or 10 because you take a=-g.

 

[velox]

are you saying that im right or wrong?

 

[Abide]

yeah, the working is fine.

it shouldn't really matter because our old teacher used to say to take the absolute value of the last result (you may end with negative time or negative distance, for example.. in which case it's essentially he same).

 

[velox]

cool thanks My teacher hasnt taught this yet.... Ic what u mean Cheers

hrmm but i still dont get the working :|

 

[xiao1985]

cool thanks My teacher hasnt taught this yet.... Ic what u mean Cheers

hrmm but i still dont get the working :|

uncouth divided the question into two halves...

first bit, he calculated the amount of time for the rock to achieve its highest moment (ie, when velocity = 0)

(v = u + at
0 = 29.4 + 9.8 * t
t = 3 s)


and then he calculated how far above the point of projection was the rock when it is at its peak...

(s = ut + ½at²
s = 0 + ½ * 9.8 * 3²
s = 44.1 m)

then he calculated the total distance the rock need to travel downward to hit the foot of the cliff:

(44.1 + 78.4 = 122.5 m)

and then the amount of time it takes for it to hit the foot of the cliff

(s = ut + ½at²
122.5 = 0 + ½ * 9.8 * t²
t = 5 s)

then add gives u 8 seconds...

 

[Abide]

yep, it may also help if you draw a diagram.

 

[velox]

now i understand Thanks guys