Finding Range, Max Height and Speed of Projectile

In summary: Speed at impact)In summary, the range of the projectile is 269.20 ft, the maximum height reached is 38.86 ft, and the speed at impact is 100 ft/s. These values were obtained using the equations for projectile motion with given initial velocity and angle of elevation.
  • #1
jheld
81
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
 
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  • #2
How did you obtain those answers?

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

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

1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It follows a curved path known as a parabola.

2. How do you find the range of a projectile?

To find the range of a projectile, you can use the formula: range = (initial velocity)^2 x sin(2 x launch angle) / gravitational acceleration. This formula assumes that air resistance is negligible.

3. How do you calculate the maximum height of a projectile?

The maximum height of a projectile can be calculated using the formula: max height = (initial velocity)^2 x (sin(launch angle))^2 / (2 x gravitational acceleration). This formula also assumes that air resistance is negligible.

4. What factors affect the speed of a projectile?

The speed of a projectile is affected by the initial velocity, launch angle, and gravitational acceleration. Air resistance can also affect the speed, but it is usually assumed to be negligible in projectile motion calculations.

5. Can you find the range and maximum height of a projectile without knowing the initial velocity?

No, the initial velocity is necessary to calculate the range and maximum height of a projectile. However, if the launch angle and gravitational acceleration are known, the initial velocity can be solved for using the equations for range and maximum height.

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