# Homework Help: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4s

1. Nov 5, 2009

### MusePhysick

1. The problem statement, all variables and given/known data
An object fired with a velocity of Vo covers a verticle distance of 20.0m and a horizontal distance of 25.0m in the 4th second of its motion. Determine the following:
a) Its velocity of projection, Vo
(the rest im not going to worry about cos if i get part a) started then i should be fine i just dont know what im doing wrong)

2. Relevant equations
ucosθ = x
usinθ = y
v = u + at2
s = ut + 1/2at2
v2 = u2 + 2as
v = s/t

3. The attempt at a solution

ok so i've tried many a times but i think i have a misconception or not seeing what is happening etc.
i know the answer as it is given which is Vo = 59.8 m/s 65.3o (angle between ground and vector)

so it takes 4 seconds for the projectile to be displaced horizontally by 25 m
so by v=s/t
v=25/4
v=6.25
which v is also known as x
so i have x=6.25 m/s

the verticle velocity i found by many ways or tried to but heres one way
s=ut + 1/2at2
s = 20
a = -9.8
t = 4
20 = 4u + 1/2.(9.8).42
98.4 = 4u
u = 24.6 m/s
and in this case u = y
so y=24.6

by pythagoras
Vo2 = y2 + x2
Vo2 = 24.62 + 6.252
Vo = 25.2815
clearly not the answer but i proceeded to find the angle
y = usinθ
θ = 14.255
angle between ground and vector = 90 - 14.255
angle = 75.745 degrees
which is clearly not the case

if you see where ive gone wrong please do tell :)
ive tried many many wasy of getting the projected velocity and nothing seemed to work, ive tried 4 times and that takes a long time

2. Nov 5, 2009

### MusePhysick

Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

sorry i used all caps for the topic, diddnt read the rules properly till it was too late, i do not wish to offend anyone soz

3. Nov 5, 2009

### Paulo Serrano

Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

I don't know, but I did the exact same thing as you.

4. Nov 5, 2009

### Pheo1986

Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

me too. theres seems to be nothing wrong with your method. i cant think of another way of doing it

5. Nov 6, 2009

### MusePhysick

Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

yeah i found out what was wrong
it says in the 4th second meaning between t=3 and t=4 so it takes 1 second for the ball to reach 20m high and 25m horizontally haha damn english, ill post the answers soon

6. Nov 6, 2009

### MusePhysick

Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

1. The problem statement, all variables and given/known data
An object fired with a velocity of Vo covers a verticle distance of 20.0m and a horizontal distance of 25.0m in the 4th second of its motion.(Meaning second 3 to second 4) Determine the following:
a) Its velocity of projection, Vo
(the rest im not going to worry about cos if i get part a) started then i should be fine i just dont know what im doing wrong)

2. Relevant equations
ucosθ = x
usinθ = y
v = u + at
s = ut + 1/2at2
v2 = u2 + 2as
v = s/t

3. The Solution

Vertical Velocity

s = ut + 1/2at2

when t=3
s1 = 3u + ½ X (-9.8) X 32

when t=4
s2 = 4u + ½ X (-9.8) X 42

s2 – s1 = 20m

s2 – s1 = (4u + ½ X (-9.8) X 42) – (3u + ½ X (-9.8) X 32)

20 = 4u – 78.4 – 3u + 44.1
20 = u – 34.3
u = 54.3 m/s
let u = y
y = 54.3 m/s

Horizontal Velocity

v = s/t
v = 25/1
v = 25 m/s
let v =x
x = 25 m/s

Vo

Vo2 = y2 + x2
Vo2 = 54.32 + 252
Vo = 59.779 m/s

tanθ = 54.3/25
θ= 65.278o

Therefore
Vo = 59.779 m/s 65.278o (angle of elevation)