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RoseWind [281]
3 years ago
13

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

ir for 4.50 s.
At what angle and with what initial speed should the ball be kicked?

Assume that the ball leaves the punker’s foot at a height of 1.23 m.
Physics
1 answer:
Flauer [41]3 years ago
6 0

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

v_x = \frac{x}{t}

v_x = \frac{67}{4.5}

v_x = 14.89m/s

The vertical component of velocity is

-h = v_y t -\frac{1}{2} gt^2

Here,

h = Height

g = Gravitational acceleration

t = Time

v_y = Vertical component of velocity

-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2

-1.23= 4.5v_y - 99.225

v_y = 21.77m/s

The direction of the velocity will be given by the tangent of the components, then

tan\theta = \frac{v_y}{v_x}

\theta = tan^{-1} (\frac{21.77}{14.89})

\theta = 55.59\°

The magnitude is given vectorially as,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{14.89^2 +21.77^2}

|V| = 26.37m/s

Therefore the angle is 55.59° and the velocity is 26.37m/s

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Explanation:

Let the speeds of father and son are v_f\ and\ v_s. The kinetic energies of father and son are K_f\ and\ K_s. The mass of father and son are  m_f\ and\ m_s

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\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}

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