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Temka [501]
3 years ago
12

A ball is launched directly in the air at a speed of 40 feet per second from a platform located 10 feet in the air. The motion o

f the ball can be modeled using the function f(t)=−16t2+40t+10, where t is the time in seconds and f(t) is the height of the ball. When, in seconds, will the ball hit the ground? Write your answer to the nearest hundredth of a second. Do not include units in your answer.
Physics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

The ball will hit the ground in 2.73 s

Explanation:

Hi there!

f(t) is the height of the ball at time "t". We need to find the value of "t" for which f(t) = 0.

f(t) = 0   Then:

-16 · t² + 40 · t + 10 = 0  

Solving the quadratic equation using the quadratic formula (a = -16, b = 40, c = 10):

t = -0.23 s and t = 2.72 s

Since time can´t be negative, be discard that value.

The ball will hit the ground in 2.73 s.

Have a nice day!

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a) F=(3675i-4543k)N

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Explanation:

a)

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r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k

Therefore it has 3 components:

r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2

We start by finding the velocity of the UFO, which is given by the derivative of the position:

v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t

And then, by differentiating again, we find the acceleration:

a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6

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F=ma

So, Substituting t = 2 s, we find:

F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N

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b)

The magnitude of a 3-dimensional vector is given by

|v|=\sqrt{v_x^2+v_y^2+v_z^2}

where

v_x,v_y,v_z are the three components of the vector

In this problem, the three components of the net force are:

F_x=3675 N\\F_y=0\\F_z=-4543 N

Therefore, substituting into the equation, we find the magnitude of the net force:

|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N

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