A ball is dropped from a 20.0 m high tower.

a. How long will it take the ball to reach the ground?

b. What is the velocity of the ball just before it hits the ground?

Answers

Answer 1

Hi there!

We can begin by using the derived kinematic equation:

[tex]t = \sqrt{\frac{2h}{g}[/tex]

Plug in the given values and let g = 9.8 m/s²:

[tex]t = \sqrt{\frac{2(20)}{g}} = \boxed{2.02 s}[/tex]

Now, we can solve for its final velocity using the equation:

[tex]v_f = v_i + at[/tex]

It is dropped from rest, so vi = 0 m/s.

[tex]v_f = at[/tex]

[tex]v_f = 9.8(2.02) = \boxed{19.796 m/s}[/tex]

Answer 2

Answer:

a. The ball would reach the ground in approximately [tex]2.02\; \rm s[/tex].

b. The velocity of the ball right before landing would be approximately [tex]19.8\; \rm m\cdot s^{-1}[/tex].

(Assumptions: the ball was dropped with no initial velocity; air resistance on the ball is negligible; [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].)

Explanation:

Under these assumptions, the acceleration of this ball would be constantly [tex]a = g = 9.81\; \rm m\cdot s^{-2}[/tex] (same as the gravitational field strength) during the descent.

Displacement of the ball: [tex]x = 20.0\; \rm m[/tex].

Initial velocity of the ball: [tex]v_{0} = 0\; \rm m\cdot s^{-1}[/tex].

Let [tex]t[/tex] denote the duration of this descent.

The SUVAT equation [tex]x = (1/2)\, a\, t^{2} + v_{0}\, t[/tex] relates the known quantities [tex]a[/tex], [tex]x[/tex], and [tex]v_{0}[/tex] to the unknown [tex]t[/tex].

Substitute the known quantities into this equation and solve to find the value of [tex]t\![/tex]:

[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2} + 0\; {\rm m \cdot s^{-1}} \times t[/tex].

[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2}[/tex].

[tex]\displaystyle t^{2} = \frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}[/tex].

Since [tex]t > 0[/tex]:

[tex]\begin{aligned}t &= \sqrt{\frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}} \\ &\approx 2.01928\; \rm s\\ &\approx 2.02\; \rm s \\ & (\text{Rounded to 2 sig. fig.})\end{aligned}[/tex].

Since the acceleration of this ball is constant, the velocity of the ball right before landing would be:

[tex]\begin{aligned}v_{1} &= a\, t \\ &\approx 9.81\; \rm m\cdot s^{-2} \times 2.01928\; \rm s \\ &\approx 19.8\; \rm m \cdot s^{-1} \end{aligned}[/tex].


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[tex] \huge \bf༆ Answer ༄[/tex]

The given terms in the question are : -

I = Electric current = 2 Amperes

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Solve for time (t)

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[tex]꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]


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#CarryOnLearning

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Hi there!

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