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astra-53 [7]
3 years ago
7

A watermelon is thrown down from a skyscraper with a speed of 7.0 m/s. It lands with an impact velocity of 20 m/s. We can ignore

air resistance. What is the displacement of the watermelon?
Physics
2 answers:
Ivenika [448]3 years ago
8 0

Answer:

17.89 m/s

Explanation:

recall that one of the equations of motion can be expressed as:

v² = u² + 2as,

where

v = final velocity = given as 20 m/s

u = initial velocity = given as 7.0 m/s

a = acceleration. Since it is freefalling downwards, the acceleration it would experience would be the acceleration due to gravity = 9.81 m/s²

s = displacement (we are asked to find)

simply substitute the known values into the equation:

v² = u² + 2as

20² = 7² + 2(9.81)s

400 - 49 = 19.62s

19.62s = 351

s = 351/19.62

s = 17.8899

s = 17.89 m/s

Volgvan3 years ago
8 0

Answer:

It’s actually -18

Explanation:

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3 years ago
gThe acceleration of gravity at the surface of Moon is 1.6 m/s2. A 5.0 kg stone thrown upward on Moon reaches a height of 20 m.
noname [10]

Answer:

(a) 8 m/s

(b) 5 s

Explanation:

(a)

Using newton's equation of motion,

v² = u²+2gs ..................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity at the surface of moon, s = height reached.

make u the subject of the equation,

u² = v²-2gs

u = √(v²-2gs)................ Equation 2

Note: As the stone is thrown up, v = 0 m/s, g is negative

Given: v = 0 m/s, s = 20 m, g = -1.6 m/s²

Substitute into equation 2

u = √(0-2×20×[-1.6])

u = √64

u = 8 m/s.

(b)

Using,

v = u+ gt

Where t = time of flight to reach the maximum height.

Make t the subject of the equation,

t = (v-u)/g................................... Equation 3

Given: v =  0 m/s, u = 8 m/s, g = - 1.6 m/s²

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t = (0-8)/-1.6

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t = 5 seconds.

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