Answer :
Answer:
Approximately [tex]10.5\; {\rm m\cdot s^{-1}}[/tex], assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that the air resistance on the ball is negligible.
Explanation:
Denote the quantities as follows:
- Let [tex]u_{x}[/tex] and [tex]u_{y}[/tex] denote the horizontal and vertical components of velocity at launch.
- Let [tex]v_{x}[/tex] and [tex]v_{y}[/tex] denote the horizontal and vertical components of velocity right before landing.
- Let [tex]u[/tex] denote the speed at launch.
- Let [tex]v[/tex] denote the speed right before landing.
- Let [tex]a_{y}[/tex] denote the vertical component of acceleration.
- Let [tex]x_{y}[/tex] denote the vertical displacement during the entire flight.
In this question, it is given that:
- Vertical displacement of the ball is [tex]x_{y} = (-17)\; {\rm m}[/tex], negative because the new position of the ball (on the ground) is below the initial position.
- Initial vertical velocity of the ball, [tex]u_{y}[/tex], is [tex]0\; {\rm m\cdot s^{-1}}[/tex] as the ball is thrown horizontally.
- Speed at landing, [tex]v[/tex], is twice the speed [tex]u[/tex] at launch. In other words, [tex]v = 2\, u[/tex].
Initial horizontal velocity [tex]u_{x}[/tex] needs to be found.
Under the assumptions:
- Horizontal velocity of the ball is constant: [tex]v_{x} = u_{x}[/tex] throughout the flight.
- Vertical acceleration of the ball would be [tex]a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex], negative because the ball is accelerating downwards.
Start by rewriting the equation [tex]v = 2\, u[/tex] in terms of the horizontal and vertical components of velocity. The speed of an object is the magnitude of its velocity. If the horizontal component of velocity is [tex]v_{x}[/tex] while the vertical component is [tex]v_{y}[/tex], the speed of the object would be:
[tex]\displaystyle v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}[/tex].
Similarly, the initial speed of this ball can be expressed in terms of the horizontal and vertical components, [tex]u_{x}[/tex] and [tex]u_{y}[/tex]:
[tex]\displaystyle u = \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex].
Therefore, the equation [tex]v = 2\, u[/tex] becomes:
[tex]\displaystyle \sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = 2\, \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex].
In this equation, [tex]v_{x} = u_{x}[/tex] under the assumption that air resistance on the ball is negligible. Therefore:
[tex]\displaystyle \sqrt{{u_{x}}^{2} + {v_{y}}^{2}} = 2\, \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex].
The next step is to express the vertical velocity [tex]v_{y}[/tex] at landing, which isn't given, in terms of vertical displacement [tex]x_{y}[/tex], vertical velocity at launch [tex]u_{y}[/tex], and acceleration [tex]a_{y}[/tex]. By the SUVAT equation that relates current velocity and initial velocity to acceleration and displacement:
[tex]\displaystyle {v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex].
[tex]{v_{y}}^{2} = {u_{y}}^{2} + 2\, a_{y}\, x_{y}[/tex].
Substitute this expression for [tex]{v_{y}}^{2}[/tex] back into the equation:
[tex]\displaystyle \sqrt{{u_{x}}^{2} + {u_{y}}^{2} + 2\, a_{y}\, x_{y}} = 2\, \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex].
Raise both sides of this equation to the power of two to eliminate the square roots:
[tex]\displaystyle {u_{x}}^{2} + {u_{y}}^{2} + 2\, a_{y}\, x_{y} = 2^{2}\, \left({u_{x}}^{2} + {u_{y}}^{2}\right)[/tex].
Rearrange this equation to isolate the term [tex]\displaystyle \left({u_{x}}^{2} + {u_{y}}^{2}\right)[/tex], which is the same as initial speed squared ([tex]u = \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex]):
[tex]\displaystyle \left({u_{x}}^{2} + {u_{y}}^{2}\right) + 2\, a_{y}\, x_{y} = 2^{2}\, \left({u_{x}}^{2} + {u_{y}}^{2}\right)[/tex]
[tex]\displaystyle 2\, a_{y}\, x_{y} = (2^{2} - 1) \, \left({u_{x}}^{2} + {u_{y}}^{2}\right)[/tex].
[tex]\displaystyle \left({u_{x}}^{2} + {u_{y}}^{2}\right) = \frac{2}{2^{2} - 1}\, a_{y}\, x_{y}[/tex].
Since [tex]u = \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex], take the square root of both sides of the equation above to obtain an expression for the initial speed [tex]u[/tex] at launch:
[tex]\displaystyle u = \sqrt{{u_{x}}^{2} + {u_{y}}^{2}} = \sqrt{\frac{2}{2^{2} - 1}\, a_{y}\, x_{y}}[/tex].
Substitute in the value of vertical acceleration [tex]a_{y} = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] and vertical displacement [tex]x_{y} = (-17)\; {\rm m}[/tex]:
[tex]\begin{aligned} u &= \sqrt{\frac{2}{2^{2} - 1}\, a_{y}\, x_{y}} \\ &= \sqrt{\frac{2}{2^{2} - 1}\, (-9.81\; {\rm m\cdot s^{-2}})\, (17\; {\rm m})} \\ &\approx 10.5\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
In other words, the speed of the ball at launch should approximately [tex]10.5\; {\rm m\cdot s^{-1}}[/tex].
