Question

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 2.20 m with an initial speed of 20.0 m/s at a downward angle of 16.0°. How much farther on the opposite floor would it have landed if the downward angle were, instead, 8.0°?

Answer 1

Well, if we solve x in this graph using some trigonometry we can know the distance for any angle.\[\tan(90-\theta)=\frac{ x }{ h } \rightarrow x=htan(90-\theta)\]If you are absolutely required to use the initial speed then you must forget the equation above and do some physics. The acceleration on the ball is gravity and nothing else (neglecting air friction). We can calculate the time of landing by using the equation of kinematics and then solving with the quadratic formula only taking into account the component of motion in the "y" direction since the components in the "x" direction do not affect the time of flight: \[h=v _{o}t+\frac{ 1 }{ 2 }at ^{2}=v _{o}^{y} t+\frac{ 1 }{ 2 }g t ^{2}\rightarrow t=\frac{ -v _{o}^{y} \pm \sqrt{(v _{o}^{y})^{2}-4(\frac{ 1 }{ 2 }g)(-h)} }{ g }\]The v initial at x term can be calculated with: \[v _{o}^{y}=v _{o}\cos \theta \rightarrow t=\frac{ -v _{o}\cos \theta\pm \sqrt{(v _{o}\cos \theta)^{2}-4(\frac{ 1 }{ 2 }g)(-h)} }{ g }\]Now once we have calculated the time of flight we can calculate the distance by multiplying the component in x of the initial velocity times the time. (There is no acceleration in the x direction)\[x=v _{o}^{x}t=v _{o}sen \theta \frac{ -v _{o}\cos \theta\pm \sqrt{(v _{o}\cos \theta)^{2}-4(\frac{ 1 }{ 2 }g)(-h)} }{ g }\]Now using that eqaution you can calculate the distance the ball travels for each angle. These are the values you need to use to compute:\[v _{o}=-22\frac{ m }{ s }\]\[g=9.8\frac{ m }{ s ^{2} }\]\[h=2.2m\]\[\theta=8°;16°\]If you want an extra challenge, you can try to figure out the difference with the distances without even computing, just by algebra knowing that one angle is half than the other and using trigonometric identities.

Answer 2

I am sorry I think you need to compute with \[h=-2.2m\]\[g=-9.8\frac{ m }{ s ^{2} }\]

Answer 3

Can somebody do the challenge?

Answer 4

Thank you very much, I really appreciate it.