Projectile Motion: Analyzing a Volleyball Jump Spike at Different Angles

[Elixer]

Homework Statement

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 18.00^o. How much farther on the opposite floor would it have landed if the downward angle were , instead, 8.00^o?
(Source - Fundamentals of PHYSICS, Halliday, Resnick,Walker, 8th edition, chapter 4, QS.33)

Homework 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 = u_yt + 0.5a_yt²
s = u_xt + 0.5a_xt²

The Attempt at a Solution

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

-2.3 = 6.18t -4.9t²
t = 1.562s

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

CASE 2:angle = 8^o
u_y = usin(8_o) = 2.78m/s

-2.3 = 2.78t-4.9t²
t = 1.025s

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

Well, I am confused.

Thank you.

 

[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.

 

[Elixer]

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

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

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

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

u_x = ucos(8^o) = 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!