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3 years ago

Anaya Goodloe decides to spike the football after a big touchdown (on Earth) and it

Physics

1 answer:

Llana [10]3 years ago

6 0

Answer:

The football leaves with the velocity, u = 15.68 m/s

Explanation:

Given data,

The football bounces back up off the ground and is airborne for, t = 3.2 s

Let the football bounces back up off the ground in the vertical direction

The formula for time of flight is given by,

t = 2u /g

∴ u = gt / 2

Substituting the values,

u = 9.8 x 3.2 / 2

u = 15.68 m/s

Hence, the football leaves with the velocity, u = 15.68 m/s

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A bullet of mass M1 is fired towards a block of mass m2 initially at rest at the edge of a frictionless table of height h as in

pantera1 [17]

The ratio of time of flight for inelastic collision to elastic collision is 1:2

The given parameters;

mass of the bullet, = m₁
mass of the block, = m₂
initial velocity of the bullet, = u₁
initial velocity of the block, = u₂

Considering inelastic collision, the final velocity of the system is calculated as;

$m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)\\\\$

The time of motion of the system form top of the table is calculated as;

$v = u + gt\\\\v = 0 + gt\\\\v = gt\\\\t= \frac{v}{g} \\\\t_A = \frac{m_1u_1}{g(m_1 + m_2)} \ \ ---(2)$

Considering elastic collision, the final velocity of the system is calculated as;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\m_1u_1 + 0 = m_1v_1 + m_2v_2\\\\m_1 u_1 = m_1v_1 + m_2v_2$

Apply one-directional velocity

$u_1 + (-v_1) = u_2 + v_2\\\\u_1 -v_1 = 0 + v_2\\\\v_1 = v_2 -u_1$

Substitute the value of $v_1$ into the above equation;

$m_1u_1 = m_1(v_2 - u_1) + m_2 v_2\\\\m_1u_1 = m_1v_2 -m_1u_1 + m_2v_2\\\\2m_1u_1 = m_1v_2 + m_2v_2\\\\2m_1u_1= v_2(m_1 + m_2)\\\\v_2 = \frac{2m_1u_1}{m_1+ m_2} \ --(3)$

where;

$v_2$ is the final velocity of the block after collision

Since the bullet bounces off, we assume that only the block fell to the ground from the table.

The time of motion of the block is calculated as follows;

$v_2 = v_0 + gt\\\\v_2 = 0 + gt\\\\t = \frac{v_2}{g} \\\\t_B = \frac{v_2}{g} \\\\ t_B = \frac{2m_1u_1}{g(m_1 + m_2)} \ \ ---(4)$

The ratio of time of flight for inelastic collision to elastic collision is calculated as follows;

$\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2$

Learn more about elastic and inelastic collision here: brainly.com/question/7694106

4 0

2 years ago

Suppose the lift force on the plane in the diagram below was larger than the force of gravity. Which of the following statements

Salsk061 [2.6K]

A because when gravity and the force are equilibrium it tend to be equal or stay at rest or constant velocity. But when the force of a plane is less than the force of gravity it will go down to the ground. Unless when the force of a plane is greater than the force of gravity, the plane will move upward at increasing rate.
Hope this helps

3 0

3 years ago

A satellite dish is in the shape of a parabolic surface. Signals coming from a satellite strike the surface of the dish and are

RSB [31]

Answer: 6.125 ft

Explanation:

If this dish has the form of a concave upward parabola and its vertex $p$ is at the origin, its corresponding equation is: