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DIA [1.3K]
3 years ago
11

A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten

tial energy ber ground level the ball leves 50% .(a) Give the total mechanical energy of the ball E in terms of maximum height Am it reaches, the mass m, and the gravitational acceleration g.(b) What is the height, hm in meters?
Physics
1 answer:
Talja [164]3 years ago
6 0

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a)  Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height,  hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

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What did j.J. Thomson discover about the composition of atoms?
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Explanation:

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3 years ago
On a level test track, a car with antilock brakes and 90% braking efficiency is determined to have a theoretical stopping distan
ser-zykov [4K]

Answer:

a = 30.832 ft/s²

Explanation:

To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0

           v² = v₀² - 2 a x

           0 = v₀² - 2ax

           a = \frac{v_o^2}{2x}

Let's start by reducing the magnitudes to ft / s              

            v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s

           

let's calculate

         a = \frac{146.66^2}{2 \ 408}

         a = 26.36 ft / s²

Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.  

Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95

                 a₁ = \frac{a_e}{0.95}

Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90

           a = \frac{a_1}{0.90}

we substitute

           a = \frac{a_e}{0.95 \ 0.90}

           

let's calculate

           a = \frac{26.36}{ 0.95 \ 0.90}

           a = 30.832 ft/s²

This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance

6 0
3 years ago
two forces have the same magnitude F, what is the angle between the two vectors if their sum has a magnitude of (a) 2F? (b) sqrt
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4 years ago
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Should be able to pull it off the same way you got it on
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