The ball rebound speed on its first bounce will be 4.79 m/s.
It should be noted that based on the information given the launch angle will be:
= 70° - 20°
= 50°
Therefore, vy/vx = tan 50
vx = vy/tan 50 = 3/t
-1.09 = vy + 0.5gt²
= 3tan50 + 4.9t²
0.95 = t²
t = ✓0.95
t = 0.97.
vx = 3/0.97
vx = 3.06m/s
vy = 3.68m/s
Therefore, the speed will be:
= ✓(3.06)² + (3.68)²
= 4.79m/s
Complete question:
A rubber ball is dropped onto a ramp that is tilted at 20° as shown in the figure(Figure 1) . A bouncing ball obeys the "law of reflection," which says that the ball leaves the surface at the same angle it approached the surface. The ball's next bounce is 3.0 m to the right of its first bounce. What is the ball rebound speed on its first bounce?
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