Question

2. A body is projected with a velocity of 20 m/s in a direction making an angle 60° with the horizontal. CCalculate (i) position after 0.5 seconds and (ii) velocity after 0.5 seconds

3. The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 m/s, what is the direction of the initial velocity? (g=9.8 m/s2 )

4. A bullet fired from a gun with a velocity of 140 m/s strikes the ground at the same level as the gun at a distance of 1 km. Find the angle of inclination with the horizontal at which the bullet is fired. (g=9.8 m/s2 )

5.A bullet is fired at an angle of 15° with the horizontal and hits the ground 6 km away. Is it possible to hit a target 10 km away by adjusting the angle of projection assuming the initial speed to be the same?

Answer 1

Answer on Question #65620-Physics-Mechanics | Relativity

2. A body is projected with a velocity of $20\,\mathrm{m/s}$ in a direction making an angle $60{}^\circ$ with the horizontal. Calculate (i) position after 0.5 seconds and (ii) velocity after 0.5 seconds

Solution

(i)

$$x = v_0 \cos \theta \quad t = 20 \cos 60 \cdot (0.5) = 5\,\mathrm{m}.$$ $$y = v_0 \sin \theta \quad t - \frac{g t^2}{2} = 20 \sin 60 \cdot (0.5) - \frac{9.8}{2} \cdot (0.5)^2 = 7.4\,\mathrm{m}.$$

(ii)

$$v_x = v_0 \cos \theta = 20 \cos 60 = 10\,\frac{\mathrm{m}}{\mathrm{s}}.$$ $$v_y = v_0 \sin \theta - g t = 20 \sin 60 - (9.8) \cdot (0.5) = 12.4\,\frac{\mathrm{m}}{\mathrm{s}}.$$

3. The maximum vertical height of a projectile is $10\,\mathrm{m}$. If the magnitude of the initial velocity is $28\,\mathrm{m/s}$, what is the direction of the initial velocity? ($g=9.8\,\mathrm{m/s}^2$)

Solution

$$h = \frac{v_0^2}{2g} \sin^2 \theta$$

The direction of the initial velocity is

$$\theta = \sin^{-1} \sqrt{\frac{2gh}{v_0^2}} = \sin^{-1} \sqrt{\frac{2(9.8)(10)}{28^2}} = 30{}^\circ$$

4. A bullet fired from a gun with a velocity of $140\,\mathrm{m/s}$ strikes the ground at the same level as the gun at a distance of $1\,\mathrm{km}$. Find the angle of inclination with the horizontal at which the bullet is fired. ($g=9.8\,\mathrm{m/s}^2$)

Solution

$$d = \frac{v_0^2}{g} \sin 2\theta$$

The angle of inclination with the horizontal is

$$\theta = \frac{1}{2} \sin^{-1} \left(\frac{g d}{v_0^2}\right) = \frac{1}{2} \sin^{-1} \left(\frac{(9.8)(1000)}{140^2}\right) = 15{}^\circ$$

5. A bullet is fired at an angle of $15{}^\circ$ with the horizontal and hits the ground $6\,\mathrm{km}$ away. Is it possible to hit a target $10\,\mathrm{km}$ away by adjusting the angle of projection assuming the initial speed to be the same?

Solution

$$d = \frac {v _ {0} ^ {2}}{g} \sin 2 \theta$$ $$\frac {d _ {1}}{d _ {2}} = \frac {\sin 2 \theta_ {1}}{\sin 2 \theta_ {2}}$$ $$\sin 2 \theta_ {2} = \frac {d _ {2}}{d _ {1}} \sin 2 \theta_ {1} = \frac {1 0}{6} \sin 2 (1 5) = \frac {5}{6} < 1$$

Yes, it is possible.

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