Projectile Motion, with only Displacement and Angle Known.

Click For Summary
SUMMARY

The discussion focuses on calculating the initial velocity of a projectile launched at a 35° angle that lands 88.5 meters away, assuming no air resistance and a gravitational acceleration of 9.80665 m/s². Participants emphasize breaking down the motion into horizontal (x) and vertical (y) components to derive equations linking initial velocity (Vi) and total time (T). The key equations utilized include the kinematic equations for projectile motion, specifically relating displacement, initial velocity, and time. The analysis concludes that while the problem presents challenges, it is solvable through systematic application of trigonometric principles and kinematic equations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic trigonometry, particularly sine and cosine functions
  • Knowledge of gravitational acceleration (9.80665 m/s²)
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to decompose vectors into components
  • Explore the effects of air resistance on projectile motion
  • Practice solving problems involving different launch angles and distances
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the dynamics of projectile motion.

n72.75
Messages
2
Reaction score
0

Homework Statement


A projectile is fired at an angle of 35° and lands 88.5 Meters Down Range.
Assume no air resistance, and a gravatational acceleration of 9.80665 M/S2

I can get as far as the trig, but it doesn't seem like there is enough information to find time or velocity.

Find Initial Velocity Of the Projectile.

Homework Equations


a = \frac{(V <sub> f </sub> - V <sub>i</sub>) }{ t }
s = V i + \frac{1}{2} at 2

The Attempt at a Solution



See Attachment.
 

Attachments

Last edited:
Physics news on Phys.org
Break up the motion into x&y components.

Since you know the distance, you can write an equation linking Vi and T (total time). Similarly because horizontal speed doesn't change, you can write an equation linking Vi and N=T/2.

Now, what happens when projectile is halfway there (with respect to the y-component of speed)?
 
Last edited:

Similar threads

Replies
40
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
23
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K