Physics, 02.04.2020 19:42, joseperez1224

A solid uniform ball rolls without slipping up a hill that is 30m tall at its peak. At the base of the hill it is
moving horizontally at 24m/s. At the top of the hill it is also moving horizontally and then it goes over
the cliff. (a) How far from the foot of the cliff does the ball land? (b) How fast is it moving when it lands?
(a) is 1 point (b) is 2 points

A solid uniform ball rolls without slipping up a hill that is 30m tall at its peak. At the base of t

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answered: barattah

(a) The horizontal distance traveled by the ball when it lands 84.28 m

(b) The speed of the ball when it lands is 34.12 m/s.

The given parameters;

height of the hill, h = 30 minitial horizontal velocity, u = 24 m/s

The final velocity of the ball is calculated by applying the principle of conservation of energy;

$\Delta K.E = \Delta P.E\\\\\frac{1}{2}m(v^2 - u^2) = mgh_1 - mgh_2 \\\\v^2 -u^2 = 2g(h_1-h_2)\\\\v^2 -u^2 = 2g (h_1-0)\\\\v^2 -u^2 = 2gh_1 \\\\v^2 = 2gh + u^2\\\\v= \sqrt{2gh + u^2} \\\\v = \sqrt{(2\times 9.8 \times 30) + 24^2} \\\\v = 34.12 \ m/s$

The time taken for the ball to reach the ground is calculated as follows;

$h = v_0_yt + \frac{1}{2} gt^2\\\\30 = 0 + \frac{1}{2} (9.8)t^2\\\\30 = 4.9t^2\\\\t^2 = \frac{30}{4.9} \\\\t^2 = 6.122\\\\t = \sqrt{6.122} \\\\t = 2.47 \ s$

The horizontal distance traveled by the ball when it lands is calculated as;

$X = v_x t\\\\X = 34.12 \times 2.47\\\\X = 84.28 \ m$

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answered: KWffrd6661

a) x = 8.75 [m]; b) v3 = 24.5[m/s]

Explanation:

To solve this problem we must divide it into two parts, the first part consists of an analysis of the conservation of energy, where energy at a certain moment must be equal to the transformation of energy at a later moment.

Ek1 = Ek2 + Ep2

where:

Ek1 = kinetic energy at point 1 [J]

Ek2 = kinetic energy at point 2 [J]

Ep2 = Potential energy at point 2 [J]

Ek1 = 0.5*m*v1^2

Ek2 = 0.5*m*v2^2

Ep2 = m*g*h

0.5*m*v1^2 = 0.5*m*v2^2 + m*g*h

Here we can eleminate the mass and determinate v2

0.5*v1^2 = 0.5*v2^2 + g*h

((0.5*v1^2 - g*h ) / 0.5 )^(1/2) = v2

replacing the values

v2 = ((0.5*(24)^2 - 9.81*30 ) / 0.5 )^(1/2)

v2 = 3.54 [m/s]

Now we can determine the time of drop of the ball using the following kinematic equation

y = vyo*t - 0.5*a*t^2

where:

y = 30[m]

vyo = 0

a = 9.81[m/s^2]

t = time[s]

-30 = -0.5*9.81*t^2

t = (30/0.5*9.81)^(1/2)

t = 2.47 [s]

And the reach can be calculated as follows.

x = 3.54 [m/s] * 2.47 [s]

x = 8.75 [m]

For speed at the end, we use the same principle of energy conservation.

0.5*m*v2^2 + m*g*h = 0.5*m*v3^2

0.5*v2^2 + g*h = 0.5*v3^2

0.5*(3.54)^2 + 9.81*(30) = 0.5*v3^2

v3 = 24.5[m/s]

answered: Guest

This did i unseated, sorry

answered: Guest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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