Question

true/false: select a for \true\ and b for \false\ mark your scantron! 17. the speed of the flowing river has no effect on the time it takes to cross a river perpendicular to its bank. 18. in the absence of air resistance, a ball thrown straight up should reach its peak exactly halfway through its flight. 19. a bullet fired perfectly horizontally would hit the ground at the same time as a bullet dropped from the same height. 20. when vectors are added head to tail, the resultant is independent of the order in which they’re added. 21. in the absence of air resistance, a bomb dropped from a plane moving in a straight line will land directly below it. 22. for an upwardly launched projectile landing on level ground, the initial vertical velocity should be the same as its final vertical velocity but in the opposite direction. problems: show work for points. write down the equation you are using to solve where possible. label units and round appropriately. 23. during wwi, a cannon called \big bertha\ was used to lob shells over the belgian border. when fired at a 40° angle, it could hit a target 12,500 m away. a. how fast would the shell be going when it left the cannon? b. what would be its maximum height? 24. george washington famously crossed the south - flowing delaware river during mid - winter. the river flows at a rate of about 1.7 m/sec and it is 180 meters wide. rowing perpendicular to the shore, it took him about 5.0 minutes (300 seconds) to cross because of ice. a. what was washington’s speed through the water (how fast can the boat go)? b. how far downstream from his launch point did the boats land? c. to an observer standing on the shore, what would have been washington’s speed relative to land?

Explanation:

Step1: Solve 23a - Range formula

The range formula for projectile motion 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 $g = 9.8\ m/s^{2}$. We need to solve for $v_{0}$. Rearranging the formula gives $v_{0}=\sqrt{\frac{Rg}{\sin2\theta}}$. Given $R = 12500\ m$ and $\theta = 40^{\circ}$, then $\sin2\theta=\sin(2\times40^{\circ})=\sin80^{\circ}\approx0.9848$. Substituting the values: $v_{0}=\sqrt{\frac{12500\times9.8}{\sin80^{\circ}}}=\sqrt{\frac{122500}{0.9848}}\approx\sqrt{124391.75}\approx352.7\ m/s$.

Step2: Solve 23b - Maximum - height formula

The formula for the maximum height of a projectile is $H=\frac{v_{0}^{2}\sin^{2}\theta}{2g}$. We know $v_{0}\approx352.7\ m/s$ and $\theta = 40^{\circ}$, so $\sin\theta=\sin40^{\circ}\approx0.6428$. Then $H=\frac{(352.7)^{2}\times(0.6428)^{2}}{2\times9.8}=\frac{124397.29\times0.4132}{19.6}=\frac{51490.85}{19.6}\approx2627\ m$.

Step3: Solve 24a - Speed formula

The width of the river $d = 180\ m$ and the time taken to cross $t = 300\ s$. The speed of the boat perpendicular to the shore (speed through the water) is given by $v=\frac{d}{t}$. Substituting the values, $v=\frac{180}{300}=0.6\ m/s$.

Step4: Solve 24b - Distance formula

The speed of the river flow $v_{r}=1.7\ m/s$ and the time taken to cross $t = 300\ s$. The distance downstream $x = v_{r}t$. Substituting the values, $x=1.7\times300 = 510\ m$.

Step5: Solve 24c - Pythagorean theorem for relative speed

The speed of the boat through the water $v_{b}=0.6\ m/s$ and the speed of the river flow $v_{r}=1.7\ m/s$. The speed relative to the land $v_{L}=\sqrt{v_{b}^{2}+v_{r}^{2}}=\sqrt{(0.6)^{2}+(1.7)^{2}}=\sqrt{0.36 + 2.89}=\sqrt{3.25}\approx1.8\ m/s$.

Answer:

23a. Approximately $352.7\ m/s$
23b. Approximately $2627\ m$
24a. $0.6\ m/s$
24b. $510\ m$
24c. Approximately $1.8\ m/s$