- #1
Eclair_de_XII
- 1,083
- 91
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°. How much farther on the opposite floor would it have landed if the downward angle were, instead, 8.00°?"
##y_0=2.3m##
##y=0m##
##x_0=0m##
##v_0=20\frac{m}{s}##
##θ=18°##
Homework Equations
##x-x_0=v_0t+\frac{1}{2}at^2##
##\vec s_0=(2.3m)j##
##\vec s=(s_x)i+(0m)j##
##\vec v_0=(19.02\frac{m}{s})i+(-6.18\frac{m}{s})j##
##\vec a=(9.8\frac{m}{s^2})j##
The Attempt at a Solution
So I've basically separated each of the positions, the initial velocity, and acceleration into four vectors. Since horizontal motion does not at all, affect vertical motion, I think I can use the usual SUVAT formulas to find the time and the distance ##s_x## using vectors. Since a ball with an initial vertical velocity of ##-6.18\frac{m}{s}## but no horizontal motion would have the same final velocity as a ball with the same vertical velocity plus horizontal motion. Yet, when I plug the numbers in, I don't quite get a sensible answer. Using the equations for only the y-parts of the factors involved...
##0m-2.3m=(-6.18\frac{m}{s})t+\frac{1}{2}(9.8\frac{m}{s^2})t^2##
##-2.3m=(-6.18\frac{m}{s})t+(4.9\frac{m}{s^2})t^2##
##4.9t^2-6.18t+2.3=0##
When I use the quadratic equation, it doesn't work out. The only way it could work out is if acceleration were negative, which I doubt.
##t=\frac{6.18±\sqrt{38.2-45.08}}{2(4.9}##
So yeah; imaginary numbers and all that. Could someone tell me what I'm doing wrong?