Three balls are tossed with same initial speed from a fourth floor dormitory window. Ball A is launched 45 degrees above the horizontal, Ball B is launched 45 degrees below the horizontal, and Ball C is launched horizontally. Which ball hits the ground with the greatest speed

Respuesta :

Answer:

All the balls hit the ground with same velocity.  

Explanation:

Let the speed of throwing be u and height of building be h.

Ball A is launched 45 degrees above the horizontal

Initial vertical speed = usin45

Vertical acceleration = -g

Height = -h

Substituting in v² = u² +2as

                       v² = (usin45)² -2 x g x (-h)

                       v² = 0.5u² +2 x g x h

Final vertical speed² = 0.5u² +2 x g x h

Initial horizontal speed = final horizontal speed =  ucos45

Final horizontal speed² = 0.5u²

Magnitude of final velocity

          [tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]  

Ball C is launched horizontally

Initial vertical speed = 0

Vertical acceleration = -g

Height = -h

Substituting in v² = 0² +2as

                       v² = 0² -2 x g x (-h)

                       v² = 2 x g x h

Final vertical speed² = 2 x g x h

Initial horizontal speed = final horizontal speed =  u

Final horizontal speed² =u²

Magnitude of final velocity

          [tex]v=\sqrt{2gh+u^2}=\sqrt{u^2+2gh}[/tex]  

Ball B is launched 45 degrees below the horizontal

Initial vertical speed = usin45

Vertical acceleration = g

Height = h

Substituting in v² = u² +2as

                       v² = (usin45)² +2 x g x h

                       v² = 0.5u² +2 x g x h

Final vertical speed² = 0.5u² +2 x g x h

Initial horizontal speed = final horizontal speed =  ucos45

Final horizontal speed² = 0.5u²

Magnitude of final velocity

          [tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]    

All the magnitudes are same so all the balls hit the ground with same velocity.  

The ball C which was launched horizontally will hit the ground fastest.

Time of motion of the balls

With respect to angle of projection above the horizontal, the time of motion each ball is calculated as follows;

[tex]T = \frac{u sin\theta }{g}[/tex]

where;

  • θ is the angle of projection above the horizontal
  • u is the initial velocity
  • T is the time to hit the ground

When θ is 45 degrees above the horizontal;

[tex]T = \frac{u \times sin(45)}{9.8}\\\\ T = 0.072u[/tex]

When θ is 45 degrees below the horizontal = 135 degrees above the horizontal

[tex]T = \frac{ u \times sin(135)}{9.8} \\\\T = 0.072 \ s[/tex]

When launched horizontally, θ = 90 degrees above the horizontal

[tex]T = \frac{u \times sin(90)}{9.8} \\\\T = 0.1 u[/tex]

Thus, the ball C which was launched horizontally will hit the ground fastest.

Learn more about time of motion here: https://brainly.com/question/2364404