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Scorpion4ik [409]
3 years ago
12

A basketball player jumps straight up with a speed of 14 m/s. How high did the player jump?

Physics
1 answer:
alexira [117]3 years ago
7 0

Answer:

<em>The basketball player jumped 10 m high.</em>

Explanation:

<u>Vertical Launch Upwards </u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=\frac{v_o^2}{2g}

The basketball player jumps up with a speed of vo=14 m/s. The maximum height is:

\displaystyle h_m=\frac{14^2}{2*9.8}

\displaystyle h_m=\frac{196}{19.6}=10

The basketball player jumped 10 m high.

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Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

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