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:

Step1: Solve for the launch - angle in the archer problem (a)

The range formula for projectile motion is $R=\frac{v^{2}\sin2\theta}{g}$, where $R$ is the range, $v$ is the initial velocity, $\theta$ is the launch - angle, and $g = 9.8\ m/s^{2}$.
We are given $R = 75.0\ m$ and $v=35.0\ m/s$. Rearranging the formula for $\sin2\theta$ gives $\sin2\theta=\frac{Rg}{v^{2}}$.
Substituting the values: $\sin2\theta=\frac{75\times9.8}{35^{2}}=\frac{735}{1225}=0.6$.
Then $2\theta=\sin^{- 1}(0.6)$, so $2\theta = 36.87^{\circ}$ or $143.13^{\circ}$. And $\theta = 18.4^{\circ}$ or $71.6^{\circ}$.

Step2: Analyze the height of the arrow at the mid - point in the archer problem (b)

The horizontal component of the initial velocity is $v_{x}=v\cos\theta$. When $\theta = 18.4^{\circ}$, $v_{x}=35\cos18.4^{\circ}\approx33.2\ m/s$.
The time to reach the mid - point ($x = 37.5\ m$) is $t=\frac{x}{v_{x}}=\frac{37.5}{33.2}\approx1.13\ s$.
The vertical component of the initial velocity is $v_{y}=v\sin\theta=35\sin18.4^{\circ}\approx11.0\ m/s$.
Using the equation $y = v_{y}t-\frac{1}{2}gt^{2}$, we substitute $v_{y}=11.0\ m/s$, $t = 1.13\ s$, and $g = 9.8\ m/s^{2}$.
$y=11\times1.13-\frac{1}{2}\times9.8\times(1.13)^{2}=12.43 - 6.24=6.19\ m$. Since $6.19\ m>3.50\ m$, the arrow will go over the branch.

Step3: Solve the standing - broad - jump problem (2)

First, find the velocity $v$ just after leaving the ground. Using $v^{2}=2ax$, where $a = 1.25g=1.25\times9.8\ m/s^{2}=12.25\ m/s^{2}$ and $x = 0.600\ m$.
$v=\sqrt{2ax}=\sqrt{2\times12.25\times0.6}=\sqrt{14.7}\approx3.83\ m/s$.
The launch - angle for maximum range in a projectile motion (assuming no air - resistance) is $\theta = 45^{\circ}$.
The range formula for projectile motion is $R=\frac{v^{2}\sin2\theta}{g}$. Since $\sin2\theta=\sin(2\times45^{\circ}) = 1$ and $v\approx3.83\ m/s$, $R=\frac{(3.83)^{2}}{9.8}=\frac{14.6689}{9.8}\approx1.497\ m$.

Step4: Solve the owl - dropping - mouse problem (3)

The vertical motion of the mouse is a free - fall. The initial vertical velocity is $v_{0y}=-3.5\sin30^{\circ}=-1.75\ m/s$ and the vertical displacement is $y=- 12.0\ m$, and $a=-g=-9.8\ m/s^{2}$.
Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, we have $-12=-1.75t-4.9t^{2}$.
Rearranging to the quadratic form $4.9t^{2}+1.75t - 12 = 0$.
The quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 4.9$, $b = 1.75$, 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 $t=\frac{-1.75 + 15.43}{9.8}=\frac{13.68}{9.8}\approx1.396\ s$.
The horizontal component of the velocity is $v_{x}=3.5\cos30^{\circ}\approx3.03\ m/s$.
The horizontal displacement is $x = v_{x}t=3.03\times1.396\approx4.23\ m$. The radius of the nest is $r = 0.15\ m$. Since the owl is $4.00\ m$ west of the nest, and the mouse travels $4.23\ m$ east, the mouse will miss the nest.

Answer:

1. (a) $\theta = 18.4^{\circ}$ or $71.6^{\circ}$; (b) The arrow will go over the branch.
2. The range of the jump is approximately $1.50\ m$.
3. The mouse will miss the nest.