You throw a ball toward a wall with a speed of 25.0 m/s and at an angle of 40.0° above the horizontal. The wall is 22.0 m horizontally from the release 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 the velocity as the ball hits the wall?

PLEASE HELP

We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of §4-5), and we let θ

0

be the angle of throw (shown in the figure). Since the horizontal component of the velocity of the ball is v

x

=v

0

cos40.0°, the time it takes for the ball to hit the wall is

t=

v

x

Δx

=

(25.0m/s)cos40.0

o

22.0m

=1.15s.

(a) The vertical distance is

Δy=(v

0

sinθ

0

)t−

2

1

gt

2

=(25.0m/s)sin40.0

o

(1.15s)−

2

1

(9.80m/s

2

)(1.15s)

2

=12.0m.

(b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: v

x

=v

0

cos40.0°=19.2m/s.

(c) The vertical component becomes (using Eq. 4-23)

v

y

=v

0

sinθ

0

−gt=(25.0m/s)sin40.0

o

−(9.80m/s

2

)(1.15s)=4.80m/s.

(d) Since v

y

>0 when the ball hits the wall, it has not reached the highest point yet.

568.98

explanation:

its math

an energy transformation flow diagram is shown

where is it ?