How do you solve this projectile motion problem?

.-----------> u = 12 m/sec
||......*
||............*
||.9.5 m.......*
||...................*
||____w______ *
||.......................
||.........................
||...........................
||............................
||.............................
||______Ground______

Let us consider the prince throws the rock horizontally at velocity u = 12 m/sec
since it has no vertical component of the velocity you can consider it as if rock was just dropped with zero velocity in vertical direction. Also as we know it hit the moat at  level 9.5 meter below, so time taken to cover that distance would be ''t" which can ce calculated using
h = ut + 1/2 gt^2         (u = 0, initial velocity in vertical direction )
9.5 = 0 + 1/2*9.81*t^2
t^2 = 19/9.81
t = 1.39 sec

Now knowing the time of flight we can calculate the width of moat as "w" = u (horizontal) * time
w = 12*1.39 = 16.70 meter   ; (initial velocity in horizontal direction u(h) = 12 m/sec)

velocity of impact
v(impact) = u (vertical) + gt
v(i) = 0 + 9.81* 1.39
v(i) = 13.636 m/sec

Hope this helped you.

Vick

Just to check. Are you saying that the initial velocity = 12 m/s AT AN ANGLE OF 42 degrees to the horizontal? Assuming that is correct, then we need to analyse the motion in the vertical and horizontal directions. I'm taking the ground level as 9.5 m below the top of the castle wall.

Vertically we can say
Vertical acceleration = - g
Vertical velocity = u sin? - gt  [? = 42; u = 12 m/s]
Vertical displacement = u sin?t - (1/2) g t^2 + 9.5

When the rock hits the ground, its vertical displacement will be zero. So we can say..

u sin?t - (1/2) g t^2 + 9.5 = 0
I'll rearrange this ....

 - (1/2) g t^2 + u sin?t + 9.5 = 0

Can you see that we now have a quadratic in t?

Using the well known formula

t = [ - u sin? ± ?( u^2 sin^2? - 4(( - (1/2)g * 9.5))] / (- g)

t = [- u sin? ± ?( u^2 sin^2? + 19g)] / (-g)

t = [- 8.03 ± ?(250.86)] / (-9.81)

t = [- 8.03 ± 15.84] / (-9.81)

t = (- 8.03 + 15.84) / (-9.81)     or     t = (-8.03 - 15.84) / (-9.81)

t = - 0.8        or     t = 2.43

Well, a negative time has no meaning so
t = 2.43 seconds.

I advise you to check my arithmetic here. You say you can manage the rest on your own? Have fun!

First a bit of physics.  As we discount drag, the only force acting on the rock while in the air is gravity.  So Ux = U cos(theta) = constant, the horizontal speed is constant throughout the flight T time.  Thus the range R = Ux T = U cos(theta) T; so that's what you need to solve to find out how far from the castle wall the rock landed (on the other side of the moat).  U = 12 mps at theta = 42 deg.

Now the T time.

y = h + Uy t - 1/2 gt^2 = (h - 9.5); where h = the launch height, y = h - 9.5 is the impact height, Uy is the initial vertical speed = U sin(theta), g is g, and t = T = ? is the time of flight you're looking for.  Theta = 42 deg re the horizontal is the launch angle.  Solve for t using a quadratic calculator, you can find good ones on the web.  I'll set it up for you.

9.5 + (12*sin(42))t - 4.9t^2 = 0 or 4.9t^2 - 8.03t - 9.5 = 0  t = T = 2.43 sec.  ANS

Then R = Ux T = 12*cos(42)*2.43 = ? meters is the moat width assuming the rock just got by it. You can do the math.

V = sqrt(Ux^2 + Vy^2) = ? the impact speed at omega = arccos(Ux/V) angle re the horizontal.  Vy^2 = 2g(H - y) = 2g(H - h + 9.5); where the apex (max height) H = h + Uy^2/2g and so H - h = Uy^2/2g.  Then combining, we have V = sqrt(Ux^2 + 2g(Uy^2/2g + 9.5)) = sqrt(Ux^2 + Uy^2 + 2g9.5) = sqrt(U^2 + 19g) = sqrt(144 + 19*9.81) = 18.18 mps.  At omega = acos(Ux/V) = acos(12*cos(42)/18.18) = 60.63 deg re the horizontal.