Question

1. an archer shoots an arrow at a 75 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?

r = \frac{v^{2}\sin2\theta}{g} g = 9.8 r = 75m v_0 = 35
r = \frac{v^{2}\sin2\theta}{g} r = 75m v = 35\cos18.45
t = \frac{37.5}{33.2}=1.13s y = v_{0y}t-\frac{1}{2}gt^{2} v_0 = 11m/s

1. 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.)
2. 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. (a)

Step1: Recall range - formula for projectile motion

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 velocity, $\theta$ is the launch - angle, and $g = 9.8\ m/s^{2}$ is the acceleration due to gravity. We are given that $R = 75\ m$ and $v_{0}=35.0\ m/s$.
So, $\sin2\theta=\frac{Rg}{v_{0}^{2}}$.

Step2: Substitute the given values

Substitute $R = 75\ m$, $v_{0}=35.0\ m/s$, and $g = 9.8\ m/s^{2}$ into the formula:
$\sin2\theta=\frac{75\times9.8}{35^{2}}=\frac{735}{1225}=0.6$.

Step3: Solve for $2\theta$ and then $\theta$

$2\theta=\sin^{- 1}(0.6)$
$2\theta\approx36.87^{\circ}$ or $143.13^{\circ}$
$\theta\approx18.4^{\circ}$ or $71.6^{\circ}$

1. (b)

Step1: Find the time to reach the halfway - point

The horizontal distance to the halfway - point is $x=\frac{R}{2}=37.5\ m$. The horizontal component of the initial velocity is $v_{0x}=v_{0}\cos\theta$. For $\theta = 18.4^{\circ}$, $v_{0x}=35\cos18.4^{\circ}\approx33.2\ m/s$. Using the formula $x = v_{0x}t$, we get $t=\frac{x}{v_{0x}}=\frac{37.5}{33.2}\approx1.13\ s$.

Step2: Find the vertical position at time $t$

The vertical component of the initial velocity is $v_{0y}=v_{0}\sin\theta$. For $\theta = 18.4^{\circ}$, $v_{0y}=35\sin18.4^{\circ}\approx11.0\ m/s$. Using the kinematic equation $y = v_{0y}t-\frac{1}{2}gt^{2}$, substitute $v_{0y}=11.0\ m/s$, $t = 1.13\ s$, and $g = 9.8\ m/s^{2}$:
$y=11\times1.13-\frac{1}{2}\times9.8\times(1.13)^{2}$
$y = 12.43-6.25=6.18\ m$. Since $6.18\ m>3.50\ m$, the arrow goes over the branch.

2.

Step1: Find the take - off velocity

Let the acceleration during the push - off be $a = 1.25g=1.25\times9.8\ m/s^{2}=12.25\ m/s^{2}$ and the distance of the push - off be $d = 0.600\ m$. Using the kinematic equation $v^{2}=v_{0}^{2}+2ad$ (where $v_{0}=0$), we get $v=\sqrt{2ad}=\sqrt{2\times12.25\times0.6}=\sqrt{14.7}\approx3.83\ m/s$.
Assume the take - off angle $\theta = 45^{\circ}$ for maximum range.

Step2: Find the range

The range formula for a projectile is $R=\frac{v^{2}\sin2\theta}{g}$. Since $\theta = 45^{\circ}$, $\sin2\theta = 1$. Substitute $v\approx3.83\ m/s$ and $g = 9.8\ m/s^{2}$ into the formula:
$R=\frac{(3.83)^{2}}{9.8}=\frac{14.6689}{9.8}\approx1.497\ m\approx1.50\ m$. Assumptions: 1. The take - off angle is $45^{\circ}$ for maximum range. 2. There is no air resistance. 3. The person's center of mass starts and ends at the same height.

3.

Answer:

Step1: Analyze the vertical motion

The vertical displacement is $y=- 12.0\ m$ (downward), the initial vertical velocity is $v_{0y}=-3.5\sin30^{\circ}=-1.75\ m/s$, and the acceleration is $a=-g=-9.8\ m/s^{2}$. Using the kinematic equation $y = v_{0y}t+\frac{1}{2}at^{2}$, we have $-12=-1.75t-4.9t^{2}$.
$4.9t^{2}+1.75t - 12 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 4.9$, $b = 1.75$, 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

The horizontal component of the velocity is $v_{0x}=3.5\cos30^{\circ}\approx3.03\ m/s$. Using the formula $x = v_{0x}t$, we get $x=3.03\times1.396\approx4.23\ m$. 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 final horizontal position of the mouse is $x_{f}=-4 + 4.23=0.23\ m>0.15\ m$. So the mouse misses the nest.