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Black_prince [1.1K]

3 years ago

6

Isaac throws an apple straight up (in the positive direction) from 1.0m above the ground, reaching a maximum height of 35 meters

. Neglecting air resistance, what is the ball's velocity when it hits the ground?

2 answers:

Assoli18 [71]3 years ago

5 0

Consider that when heigt is 35 meters, the velocity is zero. The you can use the formula for free fall with initial velocity zero

Vf = √(2gh) = √(2*9.8m/s^2 * 35m) = √686m^2/s^2 = 26.2 m/s

Assumption: 35 m is the maximum height from the ground, else, if the maximum heigth from the ground is 35 m + 1 m = 36 m, the numbers are:

Vf = √(2*9.8m/s^2*36m) = 26.6 m/s

Fittoniya [83]3 years ago

4 0

Ht = 1 + vt - 16t^2

1 + v(v/32) - 16 (v/32)^2 = 35

v = 8 Root 34

ht = 1 + 8root34 t - 16t ^2
h = o
t = 2.94

vt = 8 root34 - 32t
v = 8 root 34 - 32(2.94)
= - 47.43 ft/s

hope this helps

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$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

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