Answer:
v = 3.96 m/s
Explanation:
given,
mass of the ball = 1.2 kg
ball raised to the height = 0.80 m
ball is released from rest
velocity of the block after collision = 1 m/s
ball rebound height = 0.2 m
acceleration due to gravity = 10 m/s²
using energy conservation
[tex]mgh = \dfrac{1}{2}mv^2[/tex]
[tex]gh = \dfrac{1}{2}v^2[/tex]
[tex]9.8 \times 0.8 = \dfrac{1}{2}v^2[/tex]
[tex]v = \sqrt{2\times 9.8\times 0.8}[/tex]
v = 3.96 m/s
the speed of the ball just before the collision is equal to v = 3.96 m/s