Questionhorizontal 1.10 A stunt rider is propelled upward from his motorbike by a spring loaded ejector seat. The rider was travelling horizontally at $60 \mathrm{~km} \mathrm{~h}^{-1}$ when the ejector seat was triggered, and as they leave the seat they are travelling with a vertical velocity of $15 \mathrm{~m} \mathrm{~s}^{-1}$. The seat is 1.0 m off the ground. (a) What is the initial velocity of the stunt rider (in $\mathrm{km} \mathrm{h}^{-1}$ )? (b) How high does the stunt rider reach? (c) How far along the track does the stunt rider land on the ground? (d) What is the velocity of the stunt rider when they hit the ground (in $\mathrm{km} \mathrm{h}^{-1}$ )?
Answer: (a) $81 \mathrm{~km} \mathrm{~h}^{-1}, 42^{\circ}$ above the horizontal (b) 12 m (c) 51 m (d) $82 \mathrm{~km} \mathrm{~h}^{-1}, 43^{\circ}$ below the horizontal 1.11 A bullet is fired horizontally from a gun that is 1.5 m from the ground. The bullet travels at $1000 \mathrm{~m} \mathrm{~s}^{-1}$ and strikes a tree 150 m

Studdy Solution
حساب السرعة عند الاصطدام بالأرض.
استخدام مكونات السرعة الأفقية والرأسية النهائية. السرعة الأفقية تبقى ثابتة، والسرعة الرأسية النهائية باستخدام:
$$v_{y_{\text{final}}} = u_{y} + a t_1$$
$$v_{y_{\text{final}}} = 15 - 9.81 \times 3.12$$
$$v_{y_{\text{final}}} \approx 15 - 30.6$$
$$v_{y_{\text{final}}} \approx -15.6 \mathrm{~م} \mathrm{~ث}^{-1}$$
السرعة النهائية الكلية:
$$v_f = \sqrt{(v_{x})^2 + (v_{y_{\text{final}}})^2}$$
$$v_f = \sqrt{(16.67)^2 + (-15.6)^2}$$
$$v_f = \sqrt{278.0889 + 243.36}$$
$$v_f = \sqrt{521.4489}$$
$$v_f \approx 22.83 \mathrm{~م} \mathrm{~ث}^{-1}$$
تحويل $v_f$ إلى كم/سا:
$$v_f = 22.83 \times \frac{3600}{1000} \approx 82.19 \mathrm{~كم} \mathrm{~سا}^{-1}$$
حساب الزاوية باستخدام الدوال المثلثية:
$$\theta = \tan^{-1}\left(\frac{|v_{y_{\text{final}}}|}{v_{x}}\right)$$
$$\theta = \tan^{-1}\left(\frac{15.6}{16.67}\right)$$
$$\theta \approx 43^\circ$$
السرعة عند الاصطدام بالأرض هي $82 \mathrm{~كم} \mathrm{~سا}^{-1}, 43^\circ$ تحت الأفق.