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Crank
3 years ago
7

A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4

5 meters east of
the tower. With what magnitude was the balloon's horizontal velocity?
Physics
1 answer:
LekaFEV [45]3 years ago
4 0

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

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a model rocket flies horizontally off the edge of a cliff at a velocity of 40.0m/s. if the canyon below is 110.0m deep, how far
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189.6 m

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First of all, we study the vertical motion of the rocket in order to find the time it takes for it to land. The suvat equation for the vertical motion is

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where, taking downward as positive direction:

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Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(110)}{9.8}}=4.74 s

Now we can just analyze the horizontal motion, which is a uniform motion with constant velocity, which is equal to

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A 1300 kg steel beam is supported by two ropes. (Figure
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