How Do You Solve Projectile Motion Problems Involving a Launched Ball?

In summary, the ball launched from the edge of a table at a height of 2.5 m with an initial velocity of 5 m/s at an angle of 30 degrees above the horizontal. Using the gravitational acceleration of 10 m/s^2 and the value of cos 30 degrees as 0.86, the flight time can be calculated as 0.172 seconds and the maximum height as 25.59 m. To find the distance traveled until the ball reaches the ground, the horizontal velocity and any forces acting in the horizontal plane need to be taken into consideration.
  • #1
willydavidjr
66
0
A ball is launched from the edge of a table which is 2.5 m high. The initial velocity of the ball is 5 m/s at an angle of 30 degrees above the horizontal or table. If the gravitational acceleration is 10 m/s^2 and value of cos 30 degrees is 0.86.

a.)Calculate the flight time in seconds.
b.) Calculate the maximum height H.
c.) Calculate the distance R it travels until the ball reached the ground.


My work:

b.) For the maximum height, I first calculated the time it reaches its velocity at 0 m/s.

So V( x final)= V( x initial) cos 30 * time
But V( x final)= 0
substituting so time= 5 ( 0.86) / 10
= 0.172 seconds as the ball reaches its peak.

then y= V ( x initial) cos 30* time - 1/2 gt^2
=0.59464 + (2.5m because this is the height of the table)
= 25.59 m ( I am not sure if this is correct because I use cosine instead of sine)

I cannot get the value of flight time and the range R. Please help me.:frown:
 
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  • #2
willydavidjr said:
b.) For the maximum height, I first calculated the time it reaches its velocity at 0 m/s.

So V( x final)= V( x initial) cos 30 * time
But V( x final)= 0
substituting so time= 5 ( 0.86) / 10
= 0.172 seconds as the ball reaches its peak.
This is not correct. The final vertical velocity at maximum high will be zero. However, the final horizontal velocity will not be zero. To calcuate the flight time you need to used the equation [itex]s = u_{y}t + \frac{1}{2}a_{y}t^2[/itex], where s = -2.5 (because it finished below the table). You will obtain a quadratic equation and should solve for t. This will give you the flight time.

You should then use [itex]V_{y}^{2} = u_{y}^{2} + 2a_{y}s_{y}[/itex] to calculate the maximum height, you shouldn't use time as the trajectory is non symetrical (as it starts 2.5m up), so you cannot just halve the flight time.

As for calculating the range, ask yourself, "are there any forces acting in the horiztonal plane (if you ignore air resistance)?"

Can you go from here?
 
Last edited:
  • #3
try to always envision the velocity as the summation of two vectors, one vector in the direction of the x-axis and the other in the direction of the y-axis. this way you can break many projectile motion problems down to the core, and work out the small kinks before you put them together (with the pythagorean formula)
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air, under the influence of gravity. It follows a curved path known as a parabola.

2. What are the factors that affect projectile motion?

The factors that affect projectile motion include initial velocity, angle of launch, air resistance, and gravity. These factors can alter the path and distance traveled by the object.

3. How is the range of a projectile calculated?

The range of a projectile is calculated using the equation R = v0^2 * sin(2θ) / g, where R is the range, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

4. Can a projectile reach the same height from which it was launched?

Yes, a projectile can reach the same height from which it was launched if it is launched at a 45-degree angle and there is no air resistance. This is known as the maximum height of a projectile.

5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object, reducing its range, and altering its trajectory. It is dependent on the shape, size, and speed of the object.

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