Throwing a ball off a cliff and calculating its time, speed and using its degree.HELP

  • Thread starter MissJewels
  • Start date
  • #1

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. ;(

Homework Statement

Homework Equations

The Attempt at a Solution


Answers and Replies

  • #2

These equations usually make problems much easier:

[itex]x(t) = v_{o}cos\theta[/itex]

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

vo: Initial velocity
h: height
  • #3

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:
  • #4

Thanks for the replies, but i was wondering if someone could perhaps check my work? Tell me what im 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.
  • #5
Staff Emeritus
Science Advisor
Homework Helper

Well, you could show us your calculations after you have fixed the known errors.
  • #6

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