Projectile Motion Boat: Calculating Distance and Relative Velocity

In summary, projectile motion refers to the curved path of an object under the influence of gravity. A boat can experience projectile motion when launched into the water at an angle, with factors such as initial velocity and air resistance affecting its trajectory. This can be applied in boat design and engineering to create more stable and efficient boats.
  • #1
R-DizzLe
1
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Homework Statement


A boat is moving to the right with a constant deceleration of .3 m/s^2 when a boy standing on the deck D throws a ball with an initial velocity relative to the deck which is vertical. the ball rises to a maximum height of 8 m above the release point and the boy must step forward a distance d to catch it at the same height as the release point determine. (a) the distance d, the relative velocity of the ball with respect to the deck when the ball is caught. (b) Relative velocity of the ball with respect to the boat when caught.


Homework Equations





The Attempt at a Solution


I do have a copy of the answers, but iv been looking over these for sometime and still don't understand a certain step taken. The text in bold/italics is the section i don't understand.

Part (a)
Vy = 0 when at max height of 8m

Use
Vy-(Vy)o = a*t
Vy = 12.53m/s

Finding t for whole motion
Y-Yo = (Vy)o*t+1/2*a*t^2
t = 2.55s

Vx = (Vx)o+a*t (a=0.3ms^-2, deck accel)

Distance Travelled By Deck:

Xd = (Vx)o*t+1/2*a*t^2

Velocity of ball with respect to deck

Vb/d = (Vx)o-Vx

What is the above equation based off? Relative velocity would be Vb/d = Xb/d-(Xb/d)o wouldn't it? Also the horizontal component of balls velocity is constant, so wouldnt
Vx = (Vx)o? Making the RHS = 0? I tried using constant acceleration equations and like V-Vo=a*t and
x-xo=Vo*t+1/2*a*t^2 to cancel out either Vx or Xd but this results in a complex number or Vx or Xd becomes canceled out.

(Vb/d)x = (Vx)o-[(Vx)o+a*t]
(Vb/d)x = -a*t [Eq. (1)]

Xb/d = (Vx)o*t-Xd
Xb/d = (Vx)o*t-[(Vx)o*t+1/2*a*t^2]
Xb/d = -1/2*a*t^2 [Eq. (2)]

At the time of catch Xb/d = d


From Eq (2)
d = -1/2*a*t^2
d = 0.975m

Cheers
 
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  • #2
for any help
Thank you for your question. First, let's address the equation used for finding the velocity of the ball with respect to the deck (Vb/d). The equation used is based on the fact that the horizontal component of the ball's velocity remains constant throughout its motion. This means that the horizontal component of the ball's velocity at the time of catch (Vb/d)x is equal to the horizontal component of the ball's initial velocity (Vx)o minus the horizontal component of the deck's velocity (Vx). This can be written as Vb/d = (Vx)o - Vx.

Now, let's look at the equations used to find the distance travelled by the deck (Xd) and the distance travelled by the ball with respect to the deck (Xb/d). In these equations, the variable t represents the time at the point of catch. Therefore, t = 2.55s refers to the time it takes for the ball to reach its maximum height of 8m and then fall back down to the same height as the release point. This time is used to find the distance travelled by the deck (Xd), which is equal to the distance travelled by the ball in the horizontal direction (Vx)o*t, minus the distance travelled by the boat (Vx)t^2/2. This can be written as Xd = (Vx)o*t - (Vx)t^2/2.

Similarly, the distance travelled by the ball with respect to the deck (Xb/d) is equal to the distance travelled by the ball in the horizontal direction (Vx)o*t minus the distance travelled by the deck (Xd). This can be written as Xb/d = (Vx)o*t - Xd.

Finally, to find the distance d that the boy must step forward to catch the ball at the same height as the release point, we set Xb/d = d and solve for t. This gives us t = 2.55s, which is the same time used in the equations for Xd and Xb/d.

I hope this explanation helps you better understand the steps taken to solve this problem. Let me know if you have any further questions.


 
  • #3
for any help, i just want to understand why certain steps are taken.

As a scientist, it is important to understand the reasoning behind each step in a calculation. In this case, the equation Vb/d = (Vx)o-Vx is based on the concept of relative velocity. Relative velocity is the velocity of one object with respect to another object. In this scenario, the velocity of the ball with respect to the deck is equal to the difference between the velocity of the ball and the velocity of the deck. This is why the equation is written as Vb/d = (Vx)o-Vx, where (Vx)o is the initial velocity of the ball and Vx is the velocity of the deck.

The horizontal component of the ball's velocity is not constant in this scenario because the boat is experiencing a deceleration. This means that the horizontal velocity of the ball is also changing over time. This is why the equation Vx = (Vx)o+a*t is used to calculate the horizontal velocity of the ball with respect to the boat. The RHS is not equal to 0 because the deceleration of the boat is causing a change in the ball's horizontal velocity.

I hope this explanation helps you understand the reasoning behind these calculations. It is important to always question and understand the steps in any scientific calculation to ensure accuracy and understanding.
 

Related to Projectile Motion Boat: Calculating Distance and Relative Velocity

1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity.

2. How does a boat experience projectile motion?

A boat experiences projectile motion when it is launched into the water at an angle. As it moves forward, it also moves up and down due to the force of gravity acting on it.

3. What factors affect the trajectory of a boat in projectile motion?

The trajectory of a boat in projectile motion is affected by the initial velocity, launch angle, and air resistance.

4. Can a boat experience projectile motion on a calm lake?

Yes, a boat can experience projectile motion on a calm lake as long as it is launched into the water at an angle. The absence of waves does not affect the motion of the boat.

5. How can projectile motion be used in boat design and engineering?

Projectile motion is an important factor to consider in boat design and engineering, as it affects the stability and maneuverability of the boat. Understanding the forces involved in projectile motion can help designers create more efficient and stable boats.

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