Question

An object is launched at 40 m/s at 40 degrees from the horizontal. It lands on a platform 10 meter below the height it started from. What is delta x? (distance traveled horizontally)

Answer 1

An object is launched at $40\,\mathrm{m/s}$ at 40 degrees from the horizontal. It lands on a platform 10 meter below the height it started from. What is delta x? (distance traveled horizontally)

Solution:

$\Delta x$ – horizontal range of the object;

$V = 40\,\frac{\mathrm{m}}{\mathrm{s}}$ – initial speed of the object;

$\alpha = 40{}^{\circ}$ – the angle between the velocity and the horizontal;

$h = 10\,\mathrm{m}$ – initial height

Equation of the motion for the object, directed at angle $\alpha$: (t – time of the flight)

$V_x = V\cos \alpha; \quad V_y = V\sin \alpha;$

$$x: \Delta x = V_x t = V_t \cos \alpha$$ $$y: -h = V_t \sin \alpha - \frac{g t^2}{2}$$ $$g t^2 - 2V_t \sin \alpha - 2h = 0$$ $$9.8 t^2 - 80 t \sin 40{}^{\circ} - 20 = 0$$

Roots of a quadratic equation:

$$t_1 = 5.6\,\mathrm{s} \quad t_2 = -0.36$$

The time of the flight can be only positive: $t = 5.6\,\mathrm{s}$ (2)

(2)in(1):

$$\Delta x = V_t \cos \alpha = 40\,\frac{\mathrm{m}}{\mathrm{s}} \cdot 5.6\,\mathrm{s} \cdot \cos 40{}^{\circ} = 172\,\mathrm{m}$$

Answer: distance traveled horizontally is equal to $172\,\mathrm{m}$.