Respuesta :
Answer:
48.6 m/s
Explanation:
height, h = 0.33 m
angle of projection, θ = 3°
Let the velocity of projection is u.
Use the formula for the maximum height raised
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]
[tex]0.33=\frac{u^{2}Sin^{2}3 }{2\times 9.8}[/tex]
u = 48.6 m/s
When an object is hit at an angle it will attain the projectile motion. The ball moves with a speed of 48.6 m/sec.
What is the maximum height achieved in projectile motion?
It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion. the maximum height of motion is given by ;
h is the height = 0.33 m
θ is the angle of projection = 3°
u is the velocity of projection =?
The maximum height achieved in projectile motion is given by the formula;
[tex]\rm H=\frac{u^2sin^2\theta}{g} \\\\ \rm u = \sqrt{\frac{Hg}{sin\theta} } \\\\ \rm u = \sqrt{\frac{12.6 \times 9.81}{sin3^0}\\\\[/tex]
[tex]\rm u = 48.6 m/sec[/tex]
Hence the balls move with a speed of 48.6 m/sec.
To learn more about the projectile motion refer to the link;
https://brainly.com/question/6261898
