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Projectiles launched at an angle, neglecting air resistance

  • #1

Homework Statement



2 Problems that I am unable to solve: 1.) projectile launched at an angle 35 degrees, just clears a wall 20m high at a distance of 130m from launch. Solve for initial velocity, time it takes the projectile to reach the wall, and the velocity components + speed of the projectile when it reaches the wall.

Given: Delta X (to wall) = 130m
Theta = 35 degrees
G = -9.81m/s^2

2.) a projectile is launched at an angle of 34 degrees and travels a total distance of 230m. Solve for initial velocity and maximum height reached by projectile.

Given: Delta X = 230m
Theta = 34 degrees
Vf = 0 m/s

Homework Equations



Vx=Vi(cosine of theta) = constant

Delta X= Vi(cosine of theta)(Delta T)

Vy,f=Vi(sin of theta) - G(Delta T)

Vy,f^2 = Vi^2 (sin of theta)^2 - 2G(Delta Y)

Delta Y= Vi(sin theta)(Delta T) - 1/2G(Delta T)^2


The Attempt at a Solution



N/A; All of the equations listed call for information that is not provided ie: there are always at least two variables in the equations.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
The standard procedure for projectile motion works well on this.
Write two headings: horizontal part and vertical part
Under horizontal, you have constant speed so you write d = vt
Under vertical, you have accelerated motion, so you write
Vf = Vi + at and d = Vi*t + ½at²
Put in the numbers you know in all three equations. Use v for the initial speed, v*cos(35) for horizontal speed, v*sin(35) for vertical initial speed. Often one of the equations can be solved for the time, which is the key to the others. In this case you'll find you have two unknowns (time and initial speed) so you have to use two of the equations as a system to solve for them. Show your work here and we'll make sure you solve the problem! Remember the method for the next 2D motion problem.
 

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