An engineering student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. The cliff is 50 m above a flat, horizontal beach. How long after being released does the stone strike the beach below the cliff? what is its range?

Question

lgbasallote · Answer

why couldnt it have been the mathematician :( the engineers have led a good life :'(

anonymous · Answer

it's the author's fault

lgbasallote · Answer

okay anyway...this is motion in two dimensions, yes?

shane_b · Answer

(I bet @lgbasallote goes for a quadratic here - let's see)

lgbasallote · Answer

not actually...i havent learned this stuff yet

lgbasallote · Answer

have you forgotten im in grade school? o.O

lgbasallote · Answer

i do it by splitting it into two dimensions

lgbasallote · Answer

im take the angle with least resistance

lgbasallote · Answer

i take*

anonymous · Answer

use equations of motion to solve it....

shane_b · Answer

This one is pretty straightforward.  Start by solving for the time it takes to fall from a height of 50m:$$h=\frac{1}{2}{gt^2}$$Once you know that, multiply that times the horizontal velocity of 18m/s to get the range.

anonymous · Answer

sorry...busy answering some other questions....:|

dls · Answer

Use ==>S = Vit + ½ at ²

(- 50) = 0 +½ (-9.8)t ²
(- 50) = - ½ (9.8)t ²
(4.9)t ² = 50

t ² = 50 / 4.9
.......______
t = ²√ 10.20

t = 3.19 secs

dls · Answer

Resultant velocity is given by:
V = ²√ Vx² + Vy²

dls · Answer

The Vy here will be found even without using the t that was solved earlier, by using ==>2aS = Vf ² - Vi ²

dls · Answer

2(-9.8)(-50) = Vf ² - 0

Vf = 31.30 m/s = Vy
..........________
V = ²√ Vx² + Vy²
..........______________
V = ²√ (18.0)² + (31.30)²
..........________
V = ²√ 1303.69

V = 36.107 m/sec

dls · Answer

tanӨ = Vy / Vx
.........=(31.30) / (18)
.........=(1.74)
Ө = tan‾ ¹ (1.74)
Ө = 60.11°

dls · Answer

Find the range now?

dls · Answer

What is the answer?

shane_b · Answer

@DLS: You got the time right but I don't see how you're getting all that for the distance from a strictly horizontal throw.

dls · Answer

All these values to find the range.

shane_b · Answer

The horizontal velocity is a constant 18 m/s...until it hits the ground.

shane_b · Answer

$$(18m/s)(3.19s)=57.42m$$

dls · Answer

noooo,it is thrown at some angle?

shane_b · Answer

I don't see how there's an angle involved since it says "throws a stone horizontally over the edge "

dls · Answer

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