Answers to Mechanics Practical 4

Practical4-5.pdf

Question 1 §

$$\overrightarrow{v_0}=(130{\text{ms}}^{-1})\hat{i}+(75{\text{ms}}^{-1})\hat{j}$$

Question 1a §

Determine the maximum altitude.

\Sigma F=m\vec{a}=-mg\hat{i}\

$$\vec{a}=-g\hat{j}$$

$$a_y=-g$$

$$v_y=\int -gdt=-gt+(v_0{)}_y:(v_0{)}_y=75{\text{ms}}^{-1}$$

$$x_y=\int \left(-gt+(75{\text{ms}}^{-1})\right)dt=-\frac{1}{2}g{t}^2+(75{\text{ms}}^{-1})t+(x_0{)}_y:(x_0{)}_y=0$$

$$\text{Its maximum position is when }{x}_y^{\prime }=0:$$

$$-gt+(75{\text{ms}}^{-1})=0\therefore t=\frac{75}{9.8}\text{s}$$

$$\text{Substituting this into }{x}_y,$$

$$x_y\left(\frac{75}{9.8}\right)=-\frac{1}{2}\cdot 9.8\cdot {\left(\frac{75}{9.8}\right)}^2+(75)\left(\frac{75}{9.8}\right)$$

$$\text{Maximum altitude}=\frac{75^2}{9.8}-\frac{75^2}{2}=-\frac{219375}{98}\approx -2200\text{m}$$

Error somewhere, why is it negative?

Question 1b §

Determine the total distance travelled when the projectile first reaches the ground?

$$\text{To reach the ground, the projectile must reach its max altitude and then fall back again}$$

$$\text{This will take twice the amount of time needed to reach the max,}$$

$$t=\frac{75}{4.9}\text{s}$$

$$x_x=\int v_{x}dt=130t$$

$$\text{Total distance travelled in the x-direction}=130\cdot \frac{75}{4.9}=\frac{97500}{49}\approx 2000\text{m}$$

Question 2 §

$B:$ 1kg block

Question 1a §