Respuesta :
Answer:All balls have same velocity
Explanation:
Ball A
horizontal velocity[tex]\left ( u_x\right )=u[/tex]
vertical velocity[tex]\left ( u_y\right )=0[/tex]
let h be the height of building
Vertical velocity acquired by Ball A
[tex]u_y=\sqrt{2gh}[/tex]
Velocity just before hitting ground=[tex]\sqrt{u^2+2gh}[/tex]
Ball B
launched with velocity u at angle of 45 above horizontal
[tex]u_x=ucos45[/tex]
[tex]u_y=usin45[/tex]
horizontal velocity will remain same as there is no acceleration in that direction
vertical velocity just before hitting the floor
[tex]u_{yf}^2=\left ( usin45\right )^2+2gh[/tex]
[tex]u_{yf}=\sqrt{\frac{u^2}{2}+2gh}[/tex]
Final velocity before hitting ground [tex]v=\sqrt{u^2+2gh}[/tex]
Ball C
[tex]u_x=ucos45[/tex]
[tex]u_y=-usin45[/tex]
horizontal velocity will remain same as there is no acceleration in that direction
vertical velocity just before hitting the floor
[tex]u_{yf}^2=\left ( -usin45\right )^2+2gh[/tex]
[tex]u_{yf}=\sqrt{\frac{u^2}{2}+2gh}[/tex]
Final velocity before hitting ground [tex]v=\sqrt{u^2+2gh}[/tex]
Thus all three balls will have same final velocity.
