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

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Homework Help Overview

The problem involves a stone being thrown from a 100-meter-high cliff with an initial velocity of 25 m/s at an angle of 53° from the horizontal. The original poster seeks to determine the time before it reaches the ground, the maximum height, the horizontal reach, and the speed upon impact.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use kinematic equations to solve for time, maximum height, horizontal reach, and impact speed, but expresses confusion over the results obtained, particularly regarding the time calculation and maximum height.
  • Some participants suggest using specific equations for projectile motion and highlight the importance of ensuring the calculator is set to degrees when working with angles.
  • Others question the physical relevance of the roots obtained from the quadratic equation and suggest checking for calculation errors.

Discussion Status

The discussion is ongoing, with the original poster seeking clarification on their calculations and expressing frustration over repeated mistakes. Some participants have offered guidance on checking work and ensuring proper settings on calculators, but no consensus has been reached on the specific errors in the original poster's approach.

Contextual Notes

The original poster indicates a lack of understanding regarding the measurement of horizontal reach and the calculation of vectors, which may be contributing to their difficulties. There is also mention of potential errors in previous calculations that need to be addressed.

MissJewels
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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. ;(
 
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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.
 

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