Question

in a jump spike,a volleyball player slams the ball from overhead and toward the opposite floor .Controlling the angle of spike is difficult . Suppose a ball is spiked from a height of 2.30m with an initial speed of 20m/s at a downward angle of 18.00(*) degree .how much farther on the opposite floor would it have landed if the downward angle were,instead,8.00(*)degree?

Answer 1

I'll take a stab at this.

Step 1: Calculate the velocities in the X and Y axes.
Vi = 20.0 m/s @ 18.0 degrees
Vix = 20.0 Sin18 = 6.18 m/s
Viy = 20.0 Cos18 = 19.0 m/s

Step 2: Determine length of time for the ball to contact the floor.
d = ViyT + 1/2aT^2
2.30 = 19.0T + 1/2(9.80 m/s^2)(t^2)
0 = 4.90T^2 + 19.0T - 2.30
T = 0.122 s

Step 3: Determine distance will travel in the X direction in the same amount of time.
V = d/t => Vt = d
6.18 m/s(0.122 s) = 0.757 m

Step 4: Perform the same calculations with the angle of 8.00 degrees and compare the answer.