In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 15 m/s at an angle of 32° above the horizontal.

(a) How long does it take for the ball to reach the wall if it is 4.2 m away?
(b) How high is the ball when it hits the wall?

To answer these questions, we can use the equations of motion for projectile motion.

(a) First, let's find the time it takes for the ball to reach the wall. We will use the horizontal motion equation:

Horizontal distance = initial velocity * time * cos(angle)

Here, the horizontal distance is 4.2 m, the initial velocity is 15 m/s, the angle is 32°, and we need to find the time.

Rearranging the equation, we have:

time = horizontal distance / (initial velocity * cos(angle))

Substituting the values, we get:

time = 4.2 m / (15 m/s * cos(32°))

Using a calculator, we find that cos(32°) is approximately 0.8480. So, the equation becomes:

time = 4.2 m / (15 m/s * 0.8480)

Simplifying further, we get:

time ≈ 0.315 s

Therefore, it takes approximately 0.315 seconds for the ball to reach the wall.

(b) To find the height of the ball when it hits the wall, we can use the vertical motion equation:

Vertical displacement = initial velocity * time * sin(angle) - (1/2) * acceleration * time^2

In this case, the vertical displacement is what we need to find, the initial velocity is 15 m/s, the angle is 32°, the acceleration due to gravity is 9.8 m/s^2, and the time is 0.315 seconds (which we calculated in part (a)).

Rearranging the equation and substituting the values, we have:

Vertical displacement = (15 m/s * 0.315 s * sin(32°)) - (1/2) * (9.8 m/s^2) * (0.315 s)^2

Using a calculator, we find that sin(32°) is approximately 0.5299. So, the equation becomes:

Vertical displacement = (15 m/s * 0.315 s * 0.5299) - (1/2) * (9.8 m/s^2) * (0.315 s)^2

Simplifying further, we get:

Vertical displacement ≈ 2.637 m

Therefore, the ball is approximately 2.637 meters high when it hits the wall.