# Homework Help: Projectile Motion analyzed

1. May 13, 2010

### Elixer

1. The problem statement, all variables and given/known data
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult.Suppose a ball is spiked from a height of 2.30 m with an initial speed of 20.0 m/s at a downward angle of 18.00o. How much farther on the opposite floor would it have landed if the downward angle were , instead, 8.00o?
(Source - Fundamentals of PHYSICS, Halliday, Resnick,Walker, 8th edition, chapter 4, QS.33)

2. Relevant equations
This is a case of projectile motion , thus,
say, initial velocity = u,
displacement in vertical direction = h
displacement in horizontal direction = s
time of flight = t

h = uyt + 0.5ayt2
s = uxt + 0.5axt2

3. The attempt at a solution

CASE 1:angle = 18o
uy = usin(18o) = 6.18m/s

-2.3 = 6.18t -4.9t2
t = 1.562s

ux= ucos(18o) = 19.02 m/s
so, s1 = 19.02 * 1.562 = 29.71 m

CASE 2:angle = 8o
uy = usin(8o) = 2.78m/s

-2.3 = 2.78t-4.9t2
t = 1.025s

ux = ucos(8o) = 19.81m/s
so, s2= 19.81*1.025 = 20.31m

Well, I am confused.

Thank you.

2. May 13, 2010

### rl.bhat

-2.3 = 6.18t -4.9t^2

h, uy and ay are all in the same direction. So all should be either positive or negative.

3. May 14, 2010

### Elixer

Thank you so much! That points out the importance of signs and vector diagrams.
Yes, so the equations are:
CASE 1: angle = 18o
-2.3 = -6.18t -4.9t2
t = 0.3s

ux = ucos(18o) = 19.02m/s
so, s1 = 19.02 * 0.3 = 5.706 m

CASE 2: angle = 8o
-2.3 = -2.78t - 4.9t2
t = 0.46s

ux = ucos(8o) = 19.81m/s
so, s2 = 19.81 * 0.46 = 9.113 m

Hence , difference = s2-s1 = 9.113 - 5.706 = 3.407 m.

ANS = 3.407m

Thank you!