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BARSIC [14]
3 years ago
11

6. A quarterback can throw a receiver a high, lazy "lob" pass or a low, quick "bullet" pass.

Physics
1 answer:
BlackZzzverrR [31]3 years ago
7 0

(a) The lunch angle is 64.8⁰

(b) The initial speed of the pass when the angle of projection is 25⁰  is 21.2 m/s

(c) The time of flight of the bullet is 1.83 s

<em>"Your question is not complete, it seems to be missing the chart i uploaded".</em>

The given parameters include;

time of flight, T = 3.97 s

initial velocity, u = 21.5 m/s

(a) The lunch angle is calculated from the equation of motion of time of flight;

T = \frac{2u sin(\theta)}{g} \\\\sin(\theta ) = \frac{Tg}{2u} \\\\sin(\theta ) = \frac{3.97 \times 9.8}{2 \times 21.5} \\\\sin(\theta ) = 0.905\\\\\theta = sin^{-1} (0.905)\\\\\theta = 64.8 ^0

(b) the initial speed of the pass when the angle of projection is 25⁰ and range of 35 m, is calculated as follows;

R = \frac{u^2 sin (2\theta )}{g} \\\\u^2 = \frac{Rg}{sin(2\theta )} \\\\u = \sqrt{\frac{Rg}{sin(2\theta )}}\\\\ u = \sqrt{\frac{35 \times 9.8}{sin(2 \times 25 )} }\\\\u = 21.2 \ m/s

(c) The time of flight of the bullet is calculated as;

T = \frac{2u sin(\theta )}{g} \\\\T = \frac{2\times 21.2 \times sin(25)}{9.8} \\\\T = 1.83 \ s

Learn more here: brainly.com/question/20689870

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Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

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Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

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Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

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The initial rotational kinetic energy is given by :

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k_i=\dfrac{1}{2}\times 900\times (0.95)^2

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k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

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