# Thread: physics motion problem

1. ## physics motion problem

66. A boat travels at a speed 6.75 m/s in still water is to go directly across a river and back (Fig 3.31). The current flows 0.50 m/s (a) at what angle(s) must the boat be steered (b) How long does it take to make the roundtrip? (Assume that the boat's is constant at all times, and neglect turn around time.)

88. An astronaut on the moon fires a projectile from a launcher on a level surface so as to get the maximum range if the launcher gives the projectile a muzzle velocity 25 m/s, what is the range of the projectile [Hint: the acceleration due to gravity on the moon is only one six of that on earth.]

2. Originally Posted by cruzangyal
66. A boat travels at a speed 6.75 m/s in still water is to go directly across a river and back (Fig 3.31). The current flows 0.50 m/s (a) at what angle(s) must the boat be steered (b) How long does it take to make the roundtrip? (Assume that the boat's is constant at all times, and neglect turn around time.)
Hint: the boat's component of velocity in the direction of the flow of the water must equal the opposite of velocity of flow of the water. So if we know that the current flow is 0.50 m/s and the magnitude of the velocity of the boat is 6.75 m/s, how do you find the angle? (Sketch a diagram before you answer this.)

-Dan

3. Originally Posted by cruzangyal
88. An astronaut on the moon fires a projectile from a launcher on a level surface so as to get the maximum range if the launcher gives the projectile a muzzle velocity 25 m/s, what is the range of the projectile [Hint: the acceleration due to gravity on the moon is only one six of that on earth.]
Let the +x direction be the horizontal direction of the initial velocity (ie. parallel to the ground) and the +y direction be straight up. Set the origin of this coordinate system be at the point where the projectile is fired. Then we know that
$\displaystyle \begin{matrix}t_0 = 0~s & & t = ? \\ x_0 = 0~m & x = ? & y_0 = 0~m & y = 0~m \\ v_{0x} = 25 \cdot cos(45^o) & v_x = v_{0x} & v{0y} = 25 \cdot sin(45^o) & v_y = ? \\ a_x = 0~m/s^2 & & a_y = -g & \end{matrix}$

(I should mention that I know the angle of incline of the gun is at 45 degrees, since this provides the maximum range no matter what the g value is. I also know that $\displaystyle v_x = v_{0x}$ since $\displaystyle a_x = 0~m/s^2$. Also, y = 0 m since the projectile is being fired over a planar surface, so the projectile hits the ground at the same height it left at. g is, of course, equal to 1/6 times 9.8 m/s^2 in this case.)

Now, we want the range x. So
$\displaystyle x = x_0 + v_{0x}t$

$\displaystyle x = 25 \cdot cos(45^o) \cdot t$

We don't know t.

So look at the y information. We know that
$\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

$\displaystyle 0 = 25 \cdot sin(45^o) \cdot t - \frac{1}{12} \cdot 9.8 t^2$

Solve this for t and plug it into the x equation to get the range.

-Dan