Question

A cannon fires a shell at 200 m/s at an angle of 65 above the horizontal. a) What is the velocity of the shell after 32 s? Give both the x- and y-components, and then convert it into standard vector form, stating speed and direction. b) How long will it take for the shell to land? c) How far will the shell travel? Assume level ground.

Answer 1

Answer on Question 42937, Physics, Mechanics | Kinematics | Dynamics

a) Let the beginning of the coordinate be at initial position of a cannon. Then, the x and y coordinates of cannon are $x = v_{0}\cos \theta \cdot t$ , $y = v_{0}\sin \theta -\frac{gt^{2}}{2}$ , where $v_{0}$ is the initial velocity, $\theta$ is the angle above the horizontal. Thus, differentiating coordinates as functions of time, obtain instant velocities $v_{x} = v_{0}\cos \theta$ , $v_{y} = v_{0}\sin \theta - gt$ .

For our case $v_{0} = 200\frac{m}{s}$ , $\theta = 65$ . Hence, $v_{x}(t = 32) = 200\cos 65 \approx 84.52\frac{m}{s}$ ,

$$v _ {y} (t = 3 2) = 2 0 0 \frac {m}{s} \cdot \sin 6 5 - 9. 8 1 \frac {m}{s ^ {2}} \cdot 3 2 s = - 1 3 2. 6 6 \frac {m}{s}. \text {H e n c e}, \quad \vec {v} (t = 3 2) = (8 4. 5 2; - 1 3 2. 6 6).$$

The speed is $v(t = 32) = \sqrt{v_x^2(t = 32) + v_y^2(t = 32)} = 157.3\frac{m}{s}$ , and the angle is

$$\alpha = \arctan \left(\frac {v _ {y}}{v _ {x}}\right) \approx - 5 7. 5.$$

b) At maximum height, $v_{y} = 0$ . Hence, at this moment $t = \frac{v_{0}\sin\theta}{g}$ . The time to land is double the time to reach maximum height. Hence, $T = 2t = \frac{2v_{0}\sin\theta}{g} \approx 36.95s$ .

c) $S = v_{0}\cos \theta \cdot T\approx 3123.15m$

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