How Do You Solve Projectile Motion Problems in Physics?

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To solve projectile motion problems, it's essential to separate the motion into horizontal and vertical components. The initial vertical and horizontal velocities can be calculated using sine and cosine functions based on the launch angle. For maximum height, a different kinematic equation is needed, while the time of flight can be determined using the equations provided. It's crucial to clarify what each variable represents and ensure that the correct equations are applied to find the desired quantities. Understanding these steps will aid in solving the problem effectively.
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Homework Statement



I can't seem to understand the steps in solving these types of problems (projectile Motion), and I would like help because I have my first test on Tuesday.

A ballplayer standing at homeplate hits a baseball that is caught by another player at the same height above the ground from which it was hit. The ball is hit with an initial velocity of 22.0 m/s at an angle of 56.0° above the horizontal.

(a) How high will the ball rise?
m higher than where it was hit

(b) How much time will elapse from the time the ball leaves the bat until it reaches the fielder?
s

(c) At what distance from home plate will the fielder be when he catches the ball?
m


Homework Equations


Sin (theta) = O/H (i got 18.239)
Cos (thesta) = A/H (i got 12.302)

X= VoxT + 1/2 AT(2)
Y= VoyT + 1/2 AT(2)


The Attempt at a Solution



I used the equations to find the O and A. Then i separated into vertical and horizontal. I know that I am supposed to use these equations, but i am unsure on what I am trying to solve for. i know that A in horizontal is 0 and V in vertical is 0 (i think). I get stuck around the part where you separate the two.

Thanks in advance.
 
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bmiller13 said:

Homework Statement



I can't seem to understand the steps in solving these types of problems (projectile Motion), and I would like help because I have my first test on Tuesday.

A ballplayer standing at homeplate hits a baseball that is caught by another player at the same height above the ground from which it was hit. The ball is hit with an initial velocity of 22.0 m/s at an angle of 56.0° above the horizontal.

(a) How high will the ball rise?
m higher than where it was hit

(b) How much time will elapse from the time the ball leaves the bat until it reaches the fielder?
s

(c) At what distance from home plate will the fielder be when he catches the ball?
m


Homework Equations


Sin (theta) = O/H (i got 18.239)
Cos (thesta) = A/H (i got 12.302)

X= VoxT + 1/2 AT(2)
Y= VoyT + 1/2 AT(2)


The Attempt at a Solution



I used the equations to find the O and A. Then i separated into vertical and horizontal. I know that I am supposed to use these equations, but i am unsure on what I am trying to solve for. i know that A in horizontal is 0 and V in vertical is 0 (i think). I get stuck around the part where you separate the two.

Thanks in advance.
Your working looks good so far. The O that you found it the initial vertical velocity and the A that you found (12.302) is the initial horizontal velocity. You are also correct in assuming that the acceleration is zero in the horizontal direction and that the velocity is zero at the balls highest point. Can you know use the equations that you posted to make the next step?
 
I don't know what I am solving for in the equation, am I solving for time or X (or Y)?

If I am solving for time, then x would have to be 0 wouldn't it?
I am confused about what I would do after I solve this also.
 
bmiller13 said:
I don't know what I am solving for in the equation, am I solving for time or X (or Y)?

If I am solving for time, then x would have to be 0 wouldn't it?
I am confused about what I would do after I solve this also.
I apologise, I misread the question. For part (a) you need to use a different kinematic equation that the one you have posted and solve for X. You can then use this information to determine the flight time using the equation that you have posted and hence solve part (b).
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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