One object is thrown vertically upward with an initial velocity of 100 m/s and

1 answer:

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We have that for the Question it can be said that The maximum height reached by the first object will be 100 times that of the other.

$(H_{max})_1=100*(H_{max})_2$

From the question we are told

One object is thrown vertically upward with an initial velocity of 100 m/s and another object with an initial velocity of 10 m/s. The maximum height reached by the first object will be

that of the other.

a. 10,000 times

b. none of these

c. 1000 times

d. 100 times

e. 10 times

Generally the equation for the velocity is mathematically given as

$v=\frac{d}{t}\\\\Where\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{(V_1)^2}{(v_2)^2}\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{10000}{(10}\\\\$

$(H_{max})_1=100*(H_{max})_2$

Therefore

The maximum height reached by the first object will be 100 times that of the other.

$(H_{max})_1=100*(H_{max})_2$

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a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$