A ball is thrown straight upward and returns to the thrower's hand after 3 seconds in the air. A second ball is thrown at an angle of 30 degrees with the horizontal. At what speed (remember that this is the resultant magnitude of the vertical and horizontal speeds) must the second ball be thrown so that it reaches the same height as the one thrown vertically?

Question

contestada

Respuesta :

samuelonum1 samuelonum1 · Answer 1 · 2020-10-05T17:24:39+02:00

Explanation:

Given that the first ball is thrown vertically and upward.

since it reaches the throwers hand after 3 seconds, then the time taken to reach maximum height is

= T/2

=3/2

=1.5 seconds

Also, the projection angle for the second ball is 30°

we know that at maximum height vertical velocity is 0

therefore the final velocity of the first ball is

Vy= 0

Also, since the first ball is thrown upward the horizontal velocity is 0

Vx=0

from the first law of motion

Vy=u+at

Vy=0, a= -g

Uy=gt

Uy=9.81*1.5

Uy=14.7m/s

since both reach the same height, they will have the same vertical velocity

Uy= Vo sin∅

Vo=Uy/sin∅

Vo= 14.7/sin 30

Vo=29.4m/s