Tracking the Ball's Motion After a Throw at Ground Level

In summary, a ball is thrown into the air from ground level. After a time t = 1 s, the ball has traveled to a position x1 = 28 m to the right of and y1 = 13 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The initial velocity of the ball was v0 = 16.31 m/s and the initial angle of the throw relative to the horizontal was q0 = 50.44°. To find the height of the ball at the top of its path, you can use the equation v^2 = u^2 + 2as or the equations y = v_{y0}
  • #1
Naeem
194
0
A ball is thrown into the air from ground level. After a time t = 1 s, the ball has traveled to a position x1 = 28 m to the right of and y1 = 13 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive.



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a) What was the x component of the initial velocity of the ball?
v0x = m/s *


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b) What was the y component of the initial velocity of the ball?
v0y = m/s *
15.97 NO

HELP: Look in your book for an equation.
HELP: You need an equation relating initial velocity, the change in position, acceleration, and time.


--------------------------------------------------------------------------------
c) What was the initial speed of the throw?
v0 = m/s *

HELP: Use the answers from parts (a) and (b).


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d) What was the initial angle of the throw relative to the horizontal? Please enter your answer in degrees.
q0 = ° *


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e) What is the height of the ball at the top of its path?
h = m
8.1 NO

I need help with part e, that's it

Thanks,
 
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  • #2
Naeem said:
A ball is thrown into the air from ground level. After a time t = 1 s, the ball has traveled to a position x1 = 28 m to the right of and y1 = 13 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive.

--------------------------------------------------------------------------------
e) What is the height of the ball at the top of its path?
h = m
8.1 NO

I need help with part e, that's it
Use energy. What is the energy of the ball when thrown? When that energy is all converted into potential energy, what will its height be?

AM
 
  • #3
Not sure if using energy will work as at the top of the parabolic path horizontal component velocity would be the same as at the point of being launched so not all amount of energy at the start would be converted to P.E or am I wrong?

Try using this equation,

v^2 = u^2 + 2as

where u is vertical component velocity at launch, v is final vertical velocity, a is g and s distance traveled in the y axis.
 
  • #4
Well, I did this:

v - final velocity = 0

a = -g = - 9.81 m/s^2

so, -u^2/-2g

Can we use 'u' from part c.

do you think this would work.
 
  • #5
Naeem said:
Well, I did this:

v - final velocity = 0

a = -g = - 9.81 m/s^2

so, -u^2/-2g

Can we use 'u' from part c.

do you think this would work.
As Al pointed out, you can only use vertical velocity or its vertical kinetic energy. Its maximum height is given by:

(1) [tex]mgh = \frac{1}{2}mv_{0y}^2[/tex]

You can also use:

(2) [tex]y = v_{y0}t - \frac{1}{2}gt^2[/tex] and

(3) [tex]v_{y} = v_{y0} - gt[/tex]

and find t when v_y = 0 (ie. [itex]t = v_{y0} /g[/itex]. Substituting that value for t into (2) gives you maximum y. But you can see that it works out to the same as h in (1).

AM
 
  • #6
Hi,

You could use Andrew's method or if you'd prefer to use back the equation I wrote earlier, the u, has to be the vertical component of the initial velocity. i.e, you cannot use from (c). Use answer from (b), the y component velocity.
 

Related to Tracking the Ball's Motion After a Throw at Ground Level

1. How does the initial velocity affect the ball's motion after a throw at ground level?

The initial velocity, or the speed and direction at which the ball is thrown, is a crucial factor in determining the ball's motion after a throw. It affects both the distance and the trajectory of the ball.

2. What is the role of gravity in tracking the ball's motion after a throw at ground level?

Gravity is the force that pulls the ball towards the ground, causing it to accelerate. This acceleration affects the ball's speed and direction as it moves through the air, ultimately determining its trajectory.

3. How does air resistance impact the ball's motion after a throw at ground level?

Air resistance, or drag, is a force that acts on the ball as it moves through the air. It opposes the motion of the ball, slowing it down and changing its trajectory. The extent of air resistance depends on factors such as the shape and size of the ball, as well as the air density and speed.

4. How does the angle of the throw affect the ball's motion after a throw at ground level?

The angle of the throw, or the angle at which the ball is released, determines the initial direction of the ball's motion. A higher angle will result in a higher trajectory, while a lower angle will result in a lower trajectory. The angle also affects the distance the ball will travel.

5. How can we use equations to track the ball's motion after a throw at ground level?

There are several equations, such as the equations of motion and the projectile motion equations, that can be used to track the ball's motion after a throw at ground level. These equations take into account factors such as initial velocity, acceleration, and time to calculate the position and velocity of the ball at any given point in its trajectory.

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