Question

1. an archer shoots an arrow at a 75.0 m distant target; the bulls - eye of the target is at the same height as the release height of the arrow. (a) at what angle must the arrow be released to hit the bulls - eye if its initial speed is 35.0 m/s? in this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) there is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. will the arrow go over or under the branch?
2. in the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g. how far can they jump? state your assumptions. (increased range can be achieved by swinging the arms in the direction of the jump.)
3. an owl is carrying a mouse to the chicks in its nest. its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. the owl is flying east at 3.50 m/s at an angle 30.0° below the horizontal when it accidentally drops the mouse. is the owl lucky enough to have the mouse hit the nest? to answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

Explanation:

1.

Step1: Recall range formula for projectile

The range formula for a projectile launched and landing at the same - height is $R=\frac{v_{0}^{2}\sin2\theta}{g}$, where $R$ is the range, $v_{0}$ is the initial speed, $\theta$ is the launch angle, and $g = 9.8\ m/s^{2}$. We know $R = 75.0\ m$ and $v_{0}=35.0\ m/s$.
We need to solve for $\theta$ from $R=\frac{v_{0}^{2}\sin2\theta}{g}$. Rearranging gives $\sin2\theta=\frac{Rg}{v_{0}^{2}}$.
Substitute the values: $R = 75.0\ m$, $v_{0}=35.0\ m/s$, and $g = 9.8\ m/s^{2}$ into the formula:
$\sin2\theta=\frac{75\times9.8}{35^{2}}=\frac{735}{1225}=0.6$.

Step2: Solve for the angle

Since $\sin2\theta = 0.6$, then $2\theta=\sin^{- 1}(0.6)$. So $2\theta\approx36.9^{\circ}$ or $2\theta = 180^{\circ}-36.9^{\circ}=143.1^{\circ}$.
$\theta_{1}=\frac{36.9^{\circ}}{2}=18.45^{\circ}$ and $\theta_{2}=\frac{143.1^{\circ}}{2}=71.55^{\circ}$.

Step3: Analyze the vertical - motion to check the branch

The horizontal distance to the tree is $x = 37.5\ m$. The horizontal component of the velocity is $v_{0x}=v_{0}\cos\theta$. The time to reach the tree is $t=\frac{x}{v_{0x}}=\frac{x}{v_{0}\cos\theta}$.
Using $\theta = 18.45^{\circ}$ (we can also use $\theta = 71.55^{\circ}$), $v_{0x}=35\cos18.45^{\circ}\approx33.2\ m/s$, $t=\frac{37.5}{33.2}\approx1.13\ s$.
The vertical component of the initial velocity is $v_{0y}=v_{0}\sin\theta=35\sin18.45^{\circ}\approx11.1\ m/s$.
The vertical displacement $y = v_{0y}t-\frac{1}{2}gt^{2}$.
Substitute $v_{0y}\approx11.1\ m/s$, $t = 1.13\ s$, and $g = 9.8\ m/s^{2}$:
$y=11.1\times1.13-\frac{1}{2}\times9.8\times(1.13)^{2}=12.543-6.247 = 6.296\ m>3.50\ m$. The arrow will go over the branch.

Step1: Find the velocity at the end of the leg - extension phase

Let the acceleration $a = 1.25g=1.25\times9.8\ m/s^{2}=12.25\ m/s^{2}$ and the displacement during leg - extension $x_{1}=0.600\ m$. Using the equation $v^{2}=v_{0}^{2}+2ax$, with $v_{0} = 0\ m/s$, we get $v=\sqrt{2ax_{1}}$.
Substitute $a = 12.25\ m/s^{2}$ and $x_{1}=0.600\ m$: $v=\sqrt{2\times12.25\times0.6}=\sqrt{14.7}\approx3.83\ m/s$. This is the initial velocity for the projectile - motion part of the jump.
Assume the jumper jumps at an angle $\theta = 45^{\circ}$ for maximum range (a common assumption for projectile motion to get the maximum horizontal distance).

Step2: Use the range formula for projectile motion

The range formula for a projectile is $R=\frac{v^{2}\sin2\theta}{g}$. Since $\theta = 45^{\circ}$, $\sin2\theta=\sin90^{\circ}=1$.
We know $v\approx3.83\ m/s$ and $g = 9.8\ m/s^{2}$. So $R=\frac{(3.83)^{2}\times1}{9.8}=\frac{14.6689}{9.8}\approx1.497\ m$.

Step1: Analyze the vertical - motion of the mouse

The vertical - motion of the mouse is a free - fall motion. The initial vertical velocity is $v_{0y}=-3.5\sin30^{\circ}=-1.75\ m/s$ (negative because it is downward), the vertical displacement is $y=- 12.0\ m$ (negative because it is downward), and the acceleration is $a=-g=-9.8\ m/s^{2}$.
Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, we substitute the values: $-12=-1.75t - 4.9t^{2}$. Rearranging gives $4.9t^{2}+1.75t - 12 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $at^{2}+bt + c = 0$. Here, $a = 4.9$, $b = 1.75$, and $c=-12$.
$t=\frac{-1.75\pm\sqrt{(1.75)^{2}-4\times4.9\times(-12)}}{2\times4.9}=\frac{-1.75\pm\sqrt{3.0625 + 235.2}}{9.8}=\frac{-1.75\pm\sqrt{238.2625}}{9.8}=\frac{-1.75\pm15.43}{9.8}$.
We take the positive root $t=\frac{-1.75 + 15.43}{9.8}=\frac{13.68}{9.8}\approx1.396\ s$.

Step2: Analyze the horizontal - motion of the mouse

The horizontal component of the velocity is $v_{0x}=3.5\cos30^{\circ}\approx3.03\ m/s$.
The horizontal displacement is $x = v_{0x}t$. Substitute $v_{0x}\approx3.03\ m/s$ and $t\approx1.396\ s$: $x=3.03\times1.396\approx4.23\ m$.
The radius of the nest is $r = 0.15\ m$. The initial horizontal position of the mouse relative to the center of the nest is $x_{0}=4.00\ m$. The horizontal position of the mouse when it falls $12\ m$ is $x_{final}=4.00 - 4.23=-0.23\ m$. The magnitude of the horizontal displacement from the center of the nest is $|x_{final}| = 0.23\ m>0.15\ m$.

Answer:

(a) $\theta = 18.45^{\circ}$ or $\theta = 71.55^{\circ}$; (b) The arrow will go over the branch.

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