Question

a boy standing on the top of a tower of height 54 ft. throws a packet with a speed of 20 ft/s directly aiming towards his friend standing on the ground at a distance of 72 ft from the foot of the tower. the packet falls short of the person on the ground by x*16/3 ft. the value of x is

Answer 1

Answer on Question #44761 – Physics - Mechanics | Kinematics | Dynamics

a boy standing on the top of a tower of height 54 ft. throws a packet with a speed of 20 ft/s directly aiming towards his friend standing on the ground at a distance of 72 ft from the foot of the tower. the packet falls short of the person on the ground by $x^{*}16 / 3$ ft. the value of $x$ is

Solution:

$\mathrm{h} = 54$ ft.-height of the tower;

$\mathrm{V} = 20\frac{\mathrm{ft}}{\mathrm{s}}$ - initial velocity of the packet;

$\mathrm{d} = 72$ ft.-distance from the tower to friend;

$s = X \cdot \frac{16}{3}$ - distance from packet to friend;

$\alpha$ - angle between vertical and direction of the motion;

Components of the velocity along X-axis and Y-axis:

$\mathrm{V_x = V\cos \alpha; V_y = V\sin \alpha;}$

From the right triangle:

$\sin \alpha = \frac{\mathrm{d}}{\mathrm{AB}} = \frac{\mathrm{d}}{\sqrt{\mathrm{d}^2 + \mathrm{h}^2}} = \frac{72\mathrm{ft}}{\sqrt{(72\mathrm{ft})^2 + (54\mathrm{ft.})^2}} = \frac{4}{5}$

$$\cos \alpha = \frac {\mathrm {h}}{\mathrm {A B}} = \frac {\mathrm {h}}{\sqrt {\mathrm {d} ^ {2} + \mathrm {h} ^ {2}}} = \frac {5 4 \mathrm {f t}}{\sqrt {(7 2 \mathrm {f t}) ^ {2} + (5 4 \mathrm {f t} .) ^ {2}}} = \frac {3}{5}$$ $$\tan \alpha = \frac {\mathrm {d}}{\mathrm {h}} = \frac {7 2 \mathrm {f t .}}{5 4 \mathrm {f t .}} = \frac {4}{3}$$

Equation of motion of the particle along X-axis:

$$x: s = V _ {x} t = V t \sin \alpha$$ $$t = \frac {s}{V \sin \alpha}$$

Equation of motion of the particle along Y-axis $(g = 32\frac{ft}{s^2})$ :

$$y: h = V t \cos \alpha + \frac {g t ^ {2}}{2}$$

(1)in(2):

$$\mathrm {h} = \mathrm {V} \frac {\mathrm {s}}{\mathrm {V} \sin \alpha} \cos \alpha + \frac {\mathrm {g} \left(\frac {\mathrm {s}}{\mathrm {V} \sin \alpha}\right) ^ {2}}{2}$$ $$\mathrm {h} = \mathrm {s} \tan \alpha + \frac {\mathrm {g s} ^ {2}}{2 \mathrm {V} ^ {2} \sin^ {2} \alpha}$$ $$5 4 = \frac {4}{3} \cdot \frac {1 6}{3} \mathrm {X} + \frac {3 2 \frac {\mathrm {f t}}{\mathrm {s} ^ {2}} \cdot \left(\frac {1 6}{3} \mathrm {X}\right) ^ {2}}{2 \cdot \left(2 0 \frac {\mathrm {f t}}{\mathrm {s}}\right) ^ {2} \cdot \left(\frac {4}{5}\right) ^ {2}}$$ $$5 4 = \frac {1 6}{9} \mathrm {X} (\mathrm {X} + 4)$$ $$\frac {1 6}{9} \mathrm {X} ^ {2} + \frac {6 4}{9} \mathrm {X} - 5 4 = 0$$

Solutions of the quadratic equation:

$$X _ {1} = \frac {1}{4} (- 8 - 5 \sqrt {2 2}) \approx - 7. 8 6$$ $$X _ {2} = \frac {1}{4} (5 \sqrt {2 2} - 8) \approx 3. 8 6$$

We need only positive root, hence $X = 3.86$ .

Answer: The value X is 3.86

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