Answer:
14 m/s
Explanation:
The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:
where
v is the final speed of the stone as it reaches the water
u = 0 is the initial speed
g = 9.8 m/s^2 is the acceleration
h = 10 m is the distance covered by the stone
Solving for v, we find
Answer
given,
mass of the ball = 6.3 kg
speed of the ball = 10.4 m/s
angle made with horizontal = 43°
m_a = 1.8 kg v_a = 2.2 m/s
m_b = 1.6 kg v_b = 1.8 m/s
mass of third particle = 6.3 - 1.8 - 1.6
= 2.9 kg
u cos θ = 10.4 x cos 43° = 7.61 m/s
by using conservation momentum along x-axis
6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ
V₃ₓ = 32.44 m/s (toward right)
by using conservation momentum along y-axis
0 = 0 + 1.6 x 1.8 + 1.6 × V₃y
V₃y = -1.8 m/s (indicate downward)
velocity of the third particle
v = 32.49 m/s
θ = 3.176° (downward with horizontal)