Finding Range, Max Height and Speed of Projectile

Click For Summary

Homework Help Overview

The problem involves analyzing the motion of a projectile launched with an initial speed and angle. Participants are tasked with finding the range, maximum height, and speed at impact of the projectile.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss separating the equations of motion into x and y components. There are attempts to derive the maximum height and range using kinematic equations. Some participants question the initial calculations and the use of gravitational acceleration.

Discussion Status

The discussion includes various attempts to solve the problem, with some participants providing alternative methods and formulas for calculating range and maximum height. There is no explicit consensus on the correct answers, and multiple interpretations of the problem setup are being explored.

Contextual Notes

Participants note the importance of using the correct units for gravitational acceleration and clarify that the projectile is assumed to be launched from the origin. There is mention of discrepancies in the calculations provided by the original poster compared to expected results.

jheld
Messages
81
Reaction score
0

Homework Statement



A projectile is fired with an initial speed of 100 ft/s and angle of elevation Pi/6 (30 degrees).
Find:
A) the range of the projectile (along the x-axis)
B) the maximum height reached
C) the speed at impact


Homework Equations



Vf^2 = Vi^2 +2a*s
Vf = Vi + a*t
Sf = Si + Vo*t + (1/2)a*t^2
(where S is any coordinate axis)


The Attempt at a Solution


I have the answers from the answer key (this is a review), and I can't seem to get any of it right.
I found s (max height) (not S), to be 127.551 seconds.
I found t = 10.2 (from launch to landing)
I found range = 883.346.

I know this isn't a hard problem, but for some reason I cannot solve it.
The answers are:
A) 625*sqrt(3)/4 ft
B) 625/16 ft
C) 100 ft/s
 
Physics news on Phys.org
How did you obtain those answers?

Try separating the equations for position into the x and y components.

~Lyuokdea
 
Vf = 0 (on the way up)
0 = 100^2*sin(30)^2-2(9.8)s
solving for s gives = 127.551 ft.

for time on the way up:
127.551 = 0 + 100sin(30)t - 4.8t^2
solving for t gives = 5.1 seconds

xf = 0 + 100cos(30)(5.1)2
= 883.346

I didn't try finding the impact speed because the other two answers were wrong.
 
The problem velocity is given in units of feet per second, not meters per second. Check out your value of 'g'.
 
There's no acceleration along x-axis. The acceleration due to gravity acts only along the -y axis. 'A' implies the angle of projection.
Assuming the projectile is fired from the origin,
x=v0cosAt+0.5(0)t2
t=x/v0cosA
The net y-displacement of the projectile is zero, since the projectile returns on x-axis.
(0)=v0sinAt-0.5at2
Substituting t,
x=v02sin(2A)/g
This is the general formula for range of projectile.
Using given data,
x=269.20 ft ...(Range)

At the max height, the y-velocity of projectile is zero.
(0)=(v0sinA)2-2gy
y=(v0sinA)2/2g
This is the general formula for max height of projectile.
Using given data,
y=38.86 ft ...(Max height)

Since, there is no acceleration along x-axis,
vx=v0cosA+(0)t
The net y-displacement of projectile is zero
vy2=(v0sinA)2-2g(0)
vy=v0sinA
The components of final velocity are same as that of initial velocity, hence, the velocities must be equal. Therefore,
v=100 ft/s
 

Similar threads

Replies
40
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
26K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
2K