Intial speed of projectile motion from height

Click For Summary
SUMMARY

The discussion focuses on calculating the initial velocity of a projectile dropped from a height of 2.8 meters that travels a horizontal distance of 3 meters, without using time in the calculations. The key takeaway is that the horizontal motion has a constant velocity while the vertical motion is influenced by gravitational acceleration. The relevant kinematic equations include the relationship between vertical displacement and gravitational acceleration, specifically v^2 = v(0y)^2 - 2gy, where g is the acceleration due to gravity.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of projectile motion components
  • Familiarity with gravitational acceleration (g = 9.81 m/s²)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion
  • Learn about the effects of air resistance on projectile motion
  • Explore the concept of vector decomposition in physics
  • Investigate advanced projectile motion problems involving angles
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion dynamics without relying on time-based calculations.

ghostbuster25
Messages
99
Reaction score
0
ok I am trying to find out how to work out the intial velocity of a projectile dropped from a height h and reaches a distance s.

say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.
so can't use s=v*t

does anyone kow What is the equation for this?
 
Physics news on Phys.org
ghostbuster25 said:
ok I am trying to find out how to work out the intial velocity of a projectile dropped from a height h and reaches a distance s.

say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.
so can't use s=v*t

does anyone kow What is the equation for this?

You must know these concept. All the vectors you must consider 2 components. Horizontal Motion velocity is constant and vertical motion acceleration is constant.If the particle dropped from the hill, no need to consider angle for initial velocity.
Horizontal Motion Vertical motion
initial velocity v(0x)= v(0) v(0y)=0 (for above case)

final velocity v(x)=v(0x)=v(0) v(y)=...

acceleration a(x)=0 a(y)=-g

displacement x y

time t t (scalar)

x=v(0x)t y=v(0y)t -0.5 gt^2
v(y)=v(0y) -gt
v^2=v(0y)^2-2gy
You may use simple free fall bodies equations.
 

Similar threads

Replies
40
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 15 ·
Replies
15
Views
27K
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K