Answer:
v = 4.264 m/s
Explanation:
It is given that,
Mass of the ball, m = 0.15 kg
Spring constant of the spring, k = 150 N/m
The spring is compressed by a distance, x = 0.05 m
The pellet's initial velocity is at an angle θ (θ < 90) above the horizontal.
Let v is the final speed of the projectile when it is at a distance h = 0.800 m below its initial height. It can be calculated using the conservation of energy as :
[tex]E_i=E_f[/tex]
[tex]\dfrac{1}{2}kx^2+mgh=\dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}\times 150\times (0.05)^2+0.15\times 9.8\times 0.8=\dfrac{1}{2}\times 0.15\times v^2[/tex]
v = 4.264 m/s
So, the final speed of the projectile is 4.264 m/s. Hence, this is the required solution.
