Question

A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 13.0 m/s,at an angle of θ = 45°, from the horizontal.

The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down (as shown in the Figure). Calculate how long the ball is in the air. ( g = 9.81 m/s2)

Answer 1

nothing?

Answer 2

To find out how long the ball is in the air you first have to figure out what the initial speed of the ball is in the y direction. That is \(v_y= \sin(45) *13 m/s\).
Then you plug this into the equation
\[s = v_0t+1/2at^2\]
where \(s = 2 \) and \(v_0 = v_y\). This is a quadratic equation (as in \(ax^2+bx+c=0\) ) so you will get 2 answers. You have to figure out which one its going to be. Just remember that the way the ball goes is up and then down. Sorry, but I don't have time to do the calculation for you.

Answer 3

thank you!

Answer 4

but what would be de "a"?

Answer 5

the "a" *** excuse me

Answer 6

ok i get it

Answer 7

The a is the acceleration, which in this case would be the gravity acceleration.