Finding Range, Max Height and Speed of Projectile

  • Thread starter jheld
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  • #1
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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
 

Answers and Replies

  • #2
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How did you obtain those answers?

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

~Lyuokdea
 
  • #3
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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.
 
  • #4
PhanthomJay
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The problem velocity is given in units of feet per second, not meters per second. Check out your value of 'g'.
 
  • #5
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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 forumla 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
 

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