Ideal Projectile Motion , horizontal and vertical components

[djester555]

Homework Statement

You throw a ball toward a wall with a speed of 32 m/s at an angle of 40.0° above the horizontal directly toward a wall (Fig. 5-33). The wall is 22.0 m from the release point of the ball.


(a) How far above the release point does the ball hit the wall?
not sure what to do

(b) What are the horizontal and vertical components of its velocity as it hits the wall?

Vx = 32m/s * cos 40
Vx 24.513 i know this is correct
Vy = 32m/s * Sin 40
Vy= 20.569 this isn't correct i don't know what i did wrong

 

[rl.bhat]

Distance of the wall is given. The time taken by ball to reach the wall is
t = d/vx.
Using
y = vy*t - 1/2*g*t^2, find y, and vy at that point

 

[tomcue]

I would like to clarify that "Ideal Projectile Motion" refers to the mathematical model of motion that assumes the absence of air resistance and constant acceleration due to gravity. In this case, the ball's motion can be described by its horizontal and vertical components, with the horizontal component remaining constant and the vertical component experiencing a constant acceleration of -9.8 m/s^2.

To answer the given questions, we can use the equations of motion for projectile motion. For part (a), we can use the equation for vertical displacement:

Δy = Voy * t + 0.5 * a * t^2

Since the ball starts and ends at the same height, we can set Δy = 0. Solving for t, we get t = 4.09 s.

Next, we can use the equation for horizontal displacement:

Δx = Vx * t

Substituting the values, we get Δx = 99.02 m. Since the wall is 22.0 m away, the ball will hit the wall at a height of 77.02 m above the release point.

For part (b), the horizontal component of velocity remains constant throughout the motion, so Vx = 24.513 m/s. The vertical component changes due to acceleration, so we can use the equation:

Vy = Voy + a * t

Substituting the values, we get Vy = 20.569 m/s. However, this value is the initial vertical component of velocity, since we calculated it at the release point. To find the vertical component of velocity at the point of impact, we need to subtract the acceleration due to gravity for the time it takes to reach the wall (4.09 s). This gives us a final vertical component of velocity of 2.9 m/s.

In conclusion, the ball will hit the wall at a height of 77.02 m above the release point and with a horizontal velocity of 24.513 m/s and a vertical velocity of 2.9 m/s.