Question

A drinking fountain shoots water out from a height of 3.5cm above it's basin. The water follows a parabolic arc which extends 15cm horizontally before hitting the basin. The arc reaches a maximum height of 11cm above the basin. Determine: a) The time a single drop spends in the air. b) The initial speed of the water. and c)The angle at which it was shot out of the fountain.

Answer 1

Answer on Question #44517, Physics, Mechanics | Kinematics | Dynamics

Question:

A drinking fountain shoots water out from a height of 3.5cm above its basin. The water follows a parabolic arc which extends 15cm horizontally before hitting the basin. The arc reaches a maximum height of 11cm above the basin. Determine:

a) The time a single drop spends in the air.

b) The initial speed of the water.

c) The angle at which it was shot out of the fountain.

Answer:

a) Vertical coordinate of water drop equals:

$$y = h _ {0} + v _ {y} t - \frac {g t ^ {2}}{2}$$

Maximum height from law of conservation of energy:

$$h - h _ {0} = \frac {v _ {y} ^ {2}}{2 g}$$

Therefore:

$$v _ {y} = \sqrt {2 g (h - h _ {0})} = 1. 2 1 \frac {m}{s}$$

Now substitute $v_{y}$ into equation for $y$. Drop in air when $y > 0$. Getting quadratic equation:

$$0. 0 3 5 + 1. 2 1 t - \frac {9 . 8 t ^ {2}}{2} = 0$$

with one solution >0:

$$t = 0. 2 7 3 s$$

b) Horizontal coordinate of water drop equals:

$$x = v _ {x} t$$

Therefore:

$$v _ {x} = \frac {x}{t} = 0.549 \frac {m}{s}$$

The initial speed of the water equals:

$$v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} = 1.33 \frac {m}{s}$$

c) The tangent of angle at which drop was shot out of the fountain equals:

$$\tan \alpha = \frac {v _ {x}}{v _ {y}}$$

Therefore angle equals:

$$\alpha = \arctan 0.452 = 24.3{}^{\circ}$$

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