# Solving a Stone Thrown from a Cliff: Time, Height, Reach & Speed

• MissJewels
In summary: Thanks for the help!In summary, the stone was thrown from the cliff with an initial velocity of 25 m/s and at an angle of 53° projection from the horizontal axis. It reached the ground in 3.62 seconds.

## Homework Statement

A stone is thrown from a cliff 100 meters high with an initial velocity v0 = 25 m / s and at an angle of 53° projection from the horizontal axis. determine:
a) the time before it reaches the ground
b) the maximum height
c) its horizontal reach
d) its speed when it hits the ground

## Homework Equations

I selected these:
vx0 = v0cos(53)
vy0 = v0sin(53)

Then I applied these t the following equations:
xf= vx0t
yf= y0 + vy0t+ (1/2)(-9,8)t2

t= ( -b± √(b2 - 4ac) ) / 2a

## The Attempt at a Solution

So, for part a), using the above equations I found:

vx0 = v0cos(53)
vx0 = 25cos(53) = -22,96
xf= vx0t
xf= -22,96t

vy0 = v0sin(53)
vy0 = 25sin(53) = 9,898
yf= y0 + vy0t+ (1/2)(-9,8)t2
0= 100 + 9,898t+ (1/2)(-9,8)t2

Then, with that last equation, I used this equation: t= ( -b± √(b2 - 4ac) ) / 2a

to find t= -3,62s
When the answer SHOULD BE 7s... What am i doing wrong?

then, i tried b) using vf= 0
using the equation(H is maxheight):

vf2 = v0 -2(9,8)(H-100)
0 = 252 -16,9H +1690
H = 136,98 m
And the right answer should be 120m... GRR !

As for c), I don'T quite understand how to measure it... do i find xf?

d) Its supposed to be (15i -48,6j) m/s but i forgot how to calculate vectors.

ANYWAY PLEASE HELP! Its easy for you, but for me, this is hard. ;(

These equations usually make problems much easier:

$x(t) = v_{o}cos\theta$

$y(t) = v_{o}sin\theta - \frac{1}{2}gt^{2} + h$

vo: Initial velocity
h: height

Quadratic equations have two roots. Sometimes only one of them will make physical sense for a given situation.

EDIT: Also, make sure that your calculator is set for DEGREES rather than RADIANS if your angles are given in degrees!

Last edited:

Thanks for the replies, but i was wondering if someone could perhaps check my work? Tell me what I am doing wrong, if i miscalculated? I'm still in a nut.
Also, I have tried finding the two roots for the quadratic equation, and neither give me the proper answer.

Well, you could show us your calculations after you have fixed the known errors.

I got it! I've been repeating the same silly mistake over and over again.

## What is the formula for calculating the time a stone takes to reach the ground when thrown from a cliff?

The formula for calculating the time a stone takes to reach the ground when thrown from a cliff is t = √(2h/g), where t is the time, h is the height of the cliff, and g is the acceleration due to gravity (9.8 m/s²).

## How do you calculate the maximum height reached by a stone thrown from a cliff?

The maximum height reached by a stone thrown from a cliff can be calculated using the formula h = (v₀sinθ)²/2g, where h is the maximum height, v₀ is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

## What is the formula for determining the horizontal distance a stone reaches when thrown from a cliff?

The formula for determining the horizontal distance a stone reaches when thrown from a cliff is d = v₀cosθ * t, where d is the horizontal distance, v₀ is the initial velocity, θ is the angle of projection, and t is the time.

## How can you determine the initial velocity of a stone thrown from a cliff?

The initial velocity of a stone thrown from a cliff can be determined using the formula v₀ = √(2gh), where v₀ is the initial velocity, g is the acceleration due to gravity, and h is the height of the cliff.

## What factors can affect the speed of a stone thrown from a cliff?

The speed of a stone thrown from a cliff can be affected by factors such as the angle of projection, the initial velocity, air resistance, and the height of the cliff. Other factors such as wind and air density can also have an impact on the speed of the stone.