Respuesta :
Answer:
Range of the projectile: approximately [tex]1.06 \times 10^{3}\; {\rm m}[/tex].
Maximum height of the projectile: approximately [tex]80\; {\rm m}[/tex] (approximately [tex]45.0\; {\rm m}[/tex] above the top of the cliff.)
The projectile was in the air for approximately [tex]7.07\; {\rm s}[/tex].
The speed of the projectile would be approximately [tex]155\; {\rm m \cdot s^{-1}}[/tex] right before landing.
(Assumptions: drag is negligible, and that [tex]g = 9.81\; {\rm m\cdot s^{-1}}[/tex].)
Explanation:
If drag is negligible, the vertical acceleration of this projectile will be constantly [tex]a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. The SUVAT equations will apply.
Let [tex]\theta[/tex] denote the initial angle of elevation of this projectile.
Initial velocity of the projectile:
- vertical component: [tex]u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}[/tex]
- horizontal component: [tex]u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}[/tex].
Final vertical displacement of the projectile: [tex]x_{y} = (-35)\; {\rm m}[/tex] (the projectile landed [tex]35\: {\rm m}[/tex] below the top of the cliff.)
Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find the final vertical velocity [tex]v_{y}[/tex] of this projectile:
[tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex].
[tex]\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
(Negative since the projectile will be travelling downward towards the ground.)
Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: [tex]v_{x} = u_{x}[/tex].
The overall final velocity of this projectile will be:
[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
Change in the vertical component of the velocity of this projectile:
[tex]\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}[/tex].
Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:
[tex]\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}[/tex].
Hence, the projectile would be in the air for approximately [tex]7.07\; {\rm s}[/tex].
Also the horizontal velocity of this projectile is [tex]u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}[/tex] throughout the flight, the range of this projectile will be:
[tex]\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}[/tex].
When this projectile is at maximum height, its vertical velocity will be [tex]0[/tex]. Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find the maximum height of the projectile (relative to the top of the [tex]35\; {\rm m}[/tex] cliff.)
[tex]\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}[/tex].
Thus, the maximum height of the projectile relative to the ground will be approximately [tex]45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}[/tex].
