Finding Range, Max Height and Speed of Projectile

AI Thread Summary
The discussion focuses on calculating the range, maximum height, and speed at impact of a projectile launched at 100 ft/s at an angle of 30 degrees. The correct range is determined to be approximately 269.20 ft, while the maximum height reached is about 38.86 ft. The speed at impact remains 100 ft/s, as the horizontal and vertical components of velocity are equal upon return to the launch level. Participants emphasize the importance of separating the equations for the x and y components and correcting for gravitational acceleration. Overall, the calculations highlight common pitfalls in projectile motion problems.
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 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
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top