A ball is thrown in the air @ 30m/s

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In summary, the ball will reach a height of 45m after 3 seconds and the change in speed will be 10m/s².
  • #1
rdmfresno
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Homework Statement


A ball is thrown straight up with an initial speed of 30m/s,
(a) Show that the time it takes to reach the top of its trajectory will be 3 seconds.
(b) Show that it will reach a height of 45m (neglecting air resistance).

Homework Equations


Earth's gravity = 9.8 meter/second² (Professor said to just round it to 10).
Equation unknown.
d=1/2(g)t² ?

The Attempt at a Solution


(a) Played around with Subtracting negative -9.8m/s² (is it considered negative since the ball gets thrown up? - Instead of it coming down?).

(b) 45m = 1/2(9.8m/s²)t² - I figure I need to know "t" before I could solve this.
 
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  • #2
rdmfresno said:

Homework Statement


A ball is thrown straight up with an initial speed of 30m/s,
(a) Show that the time it takes to reach the top of its trajectory will be 3 seconds.
(b) Show that it will reach a height of 45m (neglecting air resistance).

Homework Equations


Earth's gravity = 9.8 meter/second² (Professor said to just round it to 10).
Equation unknown.

The Attempt at a Solution


(a) Played around with Subtracting negative -9.8 (is it considered negative since the ball gets thrown up? - Instead of it coming down?). I have some free-fall equations and solutions but I couldn't find the relation between them and 'throwing up in the air.'
If the speed is lowered by 10m/s² a second, how would this look in a equation?

(b) No idea where to start

Consider what the definition is of acceleration. It is the rate of change of velocity.

If something starts out at 30 m/s and slows 10 m/s every second, it will reach 0 m/s in how many seconds?

Here are some formulas for you:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

These lessons might be helpful:
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l2c.html
 
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  • #3
Thanks a bunch. Will get started on the lesson.. then try the problem again.
 

1. How high will the ball go?

Using the equation h = (v^2 sin^2θ)/2g, where h is the maximum height, v is the initial velocity (30m/s), θ is the angle of projection (typically 90 degrees for straight up), and g is the acceleration due to gravity (9.8 m/s^2), the ball will reach a maximum height of approximately 45.92 meters.

2. At what point will the ball reach its highest point?

The ball will reach its highest point when its vertical velocity becomes zero. This typically occurs when the ball reaches the peak of its trajectory, or when it has reached the maximum height.

3. How long will it take for the ball to reach its highest point?

Using the equation t = v/g, where t is the time, v is the initial velocity (30m/s), and g is the acceleration due to gravity (9.8 m/s^2), it will take approximately 3.06 seconds for the ball to reach its highest point.

4. What is the acceleration of the ball?

The acceleration of the ball is constant and is equal to the acceleration due to gravity, which is typically 9.8 m/s^2.

5. Will the ball return to its starting point?

Assuming there is no air resistance or other external factors, the ball will return to its starting point after the same amount of time it took to reach its highest point. This is due to the symmetry of the ball's trajectory, where the vertical distance traveled is the same for the ascending and descending portions of the motion.

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