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xxTIMURxx [149]

3 years ago

9

A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g

round

2 answers:

Hitman42 [59]3 years ago

8 0

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

mixas84 [53]3 years ago

6 0

Answer:0.53s

Explanation:

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amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

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$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

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substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

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suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

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<u>Answer</u>

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