Posted by
**Bobby** on
.

You throw a ball toward a wall with a speed of 25m/s and at an angleof 40 degree above the horizontal. The wall is 22m from the realease point of the ball. a) How far above the release point does the ball hit the wall? b) What are the horizontal and vertical components of its velocity as it hits the wall? c) When it hits has it passes the highest point on its trajectory?

I am confused...

The horizontal component of velocity, which remains constant is

Vox = 25 cos 40 = 19.15 m/s. The time of flight before it hits the wall is 22 m/19.15 m/s

t = 1.148 s.

The initial vertical velocity component is

Voy = 25 sin 40 = 16.07 m/s

(a) For the height y of the vertical impact point, solve

y = Voy t - (g/2) t^2

(b) The vertical velocity component at time t is

Vy(t) = Voy - gt

The horizontal component is Vox

(c) If Vy(t) <0, the ball has passed through maximum trajectory