Answer :
Answer:
Assume that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex]. Also assume that that the air resistance on this stone is negligible.
The total air time of this stone is approximately [tex]1.5\; {\rm s}[/tex].
The maximum height of this stone is [tex]1.1\; {\rm m}[/tex], same as the height of the slingshot.
The range of this stone is approximately [tex]23\; {\rm m}[/tex].
The final velocity of this stone is approximately [tex]21\; {\rm m \cdot s^{-1}}[/tex].
Explanation:
If the air resistance on this stone is negligible, this stone will accelerate towards the ground with a vertical acceleration [tex]a_{y}[/tex] of [tex]a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
At the same time, the horizontal velocity [tex]v_{x}[/tex] of this stone will stay the same during the entire flight: [tex]v_{x} = 15.2\; {\rm m\cdot s^{-1}}[/tex].
The stone was launched from a height of [tex]1.1\; {\rm m}[/tex] above the ground. Therefore, the vertical displacement [tex]x_{y}[/tex] of the stone will be [tex]x_{y} = (-1.1)\; {\rm m}[/tex] when the stone hits the ground.
The initial vertical velocity [tex]u_{y}[/tex] of this stone would be [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] since the stone was launched horizontally.
Let [tex]v_{y}[/tex] denote the final vertical velocity of this stone right before landing. Apply the SUVAT equation [tex](v_{y})^{2} - (u_{y})^{2} = 2\, a\, x[/tex] to find [tex]v_{y}\![/tex]:
[tex](v_{y})^{2} - (u_{y})^{2} = 2\, a\, x[/tex].
[tex](v_{y})^{2} = (u_{y})^{2} + 2\, a\, x[/tex].
[tex]\begin{aligned}v_{y} &= -\sqrt{(u_{y})^{2} + 2\, a\, x} \\ &= -\sqrt{0^{2} + 2\, (-9.81) \, (-1.1)}\; {\rm m\cdot s^{-1}} \\ &\approx -14.6908 \end{aligned}[/tex].
(Final vertical velocity [tex]v_{y}[/tex] is negative since the stone is travelling downwards toward the ground.)
Let [tex]t[/tex] denote the duration of this flight. Apply the SUVAT equation [tex]t = (v_{y} - u_{y}) / (a)[/tex] to find [tex]t\![/tex]:
[tex]\begin{aligned} t &= \frac{v_{y} - u_{y}}{a} \\ &\approx \frac{(-14.6908) - 0}{(-9.81)}\; {\rm s} \\ &\approx 1.49753\; {\rm s} \end{aligned}[/tex].
In other words, this stone was in the air for approximately [tex]1.5\; {\rm s}[/tex].
Also because the stone was launched horizontally, the vertical velocity of this stone started at [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] and became negative (toward the ground) immediately after.
Hence, this stone would not have travelled upward during the entire flight. The height of this stone would be maximized immediately after the stone was launched: [tex]1.1\; {\rm m}[/tex].
Multiple the horizontal velocity [tex]v_{x} = 15.2\; {\rm m\cdot s^{-1}}[/tex] of this stone by the duration of the flight [tex]t \approx 1.49753\; {\rm s}[/tex] to find the range (horizontal displacement) of this stone:
[tex]\begin{aligned} (\text{range}) &= v_{x}\, t \\ &\approx (15.2\; {\rm m\cdot s^{-1}})\, (1.49753\; {\rm s}) \\ &\approx 23\; {\rm m} \end{aligned}[/tex].
Right before landing, the stone would be travelling with a vertical velocity of [tex]v_{y} \approx (-14.6908)\; {\rm m\cdot s^{-1}}[/tex] and a horizontal velocity of [tex]v_{x} = 15.2\; {\rm m\cdot s^{-1}}[/tex]. Apply the Pythagorean Theorem to find the overall velocity of this stone at that moment:
[tex]\begin{aligned} v &= \sqrt{{(v_{x})}^{2}\, {(v_{y})^{2}} \\ &\approx \sqrt{(15.2\; {\rm m\cdot s^{-1})^{2} + (-14.6908)^{2}}\; {\rm m\cdot s^{-1}} \\ &\approx 21\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
