- #1

itsme24

- 8

- 0

Here is the problem:

A projectile of mass 19.4 kg is fired at an angle of 57.0 degrees above the horizontal and with a speed of 81.0 m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance.

And here is what I did:

I first found the center of mass when y = 0 since the center of mass follows the trajectory of the parabola I used the equation:

y(x) = tan(theta)x - .5g(x/(V_i*cos(theta))^2

y(x) = 0

theta = 57 degrees

g= 9.8m/s^2

V_i = 81.0 m/s so...

0 = tan(57)x - 4.9(x/(8.1*cos(57))^2

0 = 1.54x - 0.00252x^2

quadratic:

x = 0,

**776.1m = center of mass**

Then it says that the first fragment drops vertically at the highest point and this is partly where I get confused since with my math I assumed that means straight down, correct me if I'm wrong. So I found t when V_y = 0 to find x at that point.

V_y = 0 = V_yi + a_y*t

0= 81.0*sin(57) - 9.8t

**t= 6.93s**

x when t = 6.93s which should give me the position fragment 1 landed:

x_f1 = x_i + V_ix*t + 0.5a_x*t^2

x_f1 = 0 + 81.0cos(57) + 0

**x_f1 = 305.72m**

Then I just had to find x of fragment 2:

x_cm = (m_f1*x_f1) + (m_f2*x_f2) / (m_f1 + m_f2)

so, I made the center of mass the 0 coordinate:

776.1m = [(9.7kg*-470.4m) + (9.7kg*x_f2)] / (9.7kg*2)

**x_f2 = 2020m**, which turned out to be wrong!

I'm sorry this problem is so long