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

AI Thread Summary
A stone is thrown from a 100-meter-high cliff at an initial velocity of 25 m/s and an angle of 53°. The discussion focuses on calculating the time to reach the ground, maximum height, horizontal reach, and impact speed using projectile motion equations. The user struggles with finding the correct time of flight and maximum height, initially obtaining incorrect values due to calculation errors and confusion with quadratic roots. They also emphasize the importance of ensuring the calculator is set to degrees for angle measurements. Ultimately, the user realizes they have been making repeated mistakes in their calculations.
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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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