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

[willydavidjr]

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.

 

[Hootenanny]

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 s = u_{y}t + \frac{1}{2}a_{y}t^2, 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 V_{y}^{2} = u_{y}^{2} + 2a_{y}s_{y} 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?

 

[93157]

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)