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3 years ago

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A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above th

e ground, how long will it take him to reach a net that is 8 ft above the ground

1 answer:

jok3333 [9.3K]3 years ago

6 0

<h2>It will take 0.125 seconds to reach the net.</h2>

Explanation:

Initial speed, u = 34 ft/s = 10.36 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m

We have equation of motion, s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

1.22 = 10.36 x t + 0.5 x -9.81 x t²

4.905t² - 10.36 t + 1.22 = 0

t = 1.99 s or t = 0.125 seconds

Minimum time is 0.125 seconds.

It will take 0.125 seconds to reach the net.

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To understand how to find the velocities of objects after a collision.

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There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

<u>Part A</u>: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$

$v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m.v}{m+m}$

v_1 = v_2 = $\frac{v}{2}$

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = $\frac{2m - m}{2m + m } . v$

v_1 = $\frac{v}{3}$

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

v_1 = v_2 = $\frac{v}{4}$

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