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Firdavs [7]
4 years ago
5

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 5.00 m above

the ground and measure that it hits the ground 0.804 s later. Part APart complete What is the acceleration of gravity near the surface of this planet
Physics
1 answer:
GenaCL600 [577]4 years ago
4 0

Answer:

g = 15.5 m/s²

Explanation:

In order to find the acceleration due to gravity near the surface of this planet can be calculated by using 2nd equation of motion. The 2nd equation of motion is given as:

h = Vi t + (0.5)gt²

where,

h = height covered by the wrench = 5 m

Vi = Initial Velocity = 0 m/s

t = Time Taken to hit the ground = 0.804 s

g = acceleration due to gravity near the surface of the planet = ?

Therefore,

5 m = (0 m/s)(0.804 s) + (0.5)(g)(0.804 s)²

g = (5 m)/(0.3232 s²)

<u>g = 15.5 m/s²</u>

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a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle
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31.035^{\circ}

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x = Displacement in x direction = 5.34 m

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Displacement in x direction is given by

x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}

Displacement in y direction is given by

y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}

\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}

The angle at which the ball was thrown is 31.035^{\circ}.

7 0
3 years ago
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