Physics, 04.01.2021 14:20, shakira11harvey6

A projectile is shot a cliff of 20m high, at an angle of 60o with respect to the horizontal, and it lands on the ground 8 seconds later. Find: a) the initial speed ? b) the speed of the projectile after 4 s c) the horizontal range d)the maximum height it reached e) what was the speed of the projectile when it landed on the ground?

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answered: frankly106

a) The initial speed of the projectile is approximately 42.410 meters per second.

b) The speed of the projectile after 4 seconds is approximately 21.352 meters per second.

c) The horizontal range of the projectile is 169.64 meters.

d) The maximum height of the projectile is 68.775 meters.

e) The speed of the projectile when it landed on the ground is approximately 46.807 meters per second.

Explanation:

According to the statement, the projects shows a parabolic motion, which consists in the combination of horizontal uniform motion and vertical uniformly accelerated motion due to gravity. This motion is represented by the following equations of motion:

$x = x_{o} + v_{o}\cdot t\cdot \cos \theta$ (1)

$y = y_{o} + v_{o}\cdot t\cdot \sin \theta + \frac{1}{2}\cdot g\cdot t^{2}$ (2)

Where:

$x_{o}$ , - Initial and current horizontal position, measured in meters.

$y_{o}$ , - Initial and current vertical position, measured in meters.

$\theta$ - Launch angle, measured in sexagesimal angle.

- Gravitational acceleration, measured in meters per square second.

- Time, measured in second.

a) By using (2) and knowing that $y_{o} = 20\,m$ , $y = 0\,m$ , $t = 8\,s$ , $\theta = 60^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then initial speed of the projectile is:

$v_{o}\cdot t \cdot \sin \theta = y-y_{o}-\frac{1}{2}\cdot g\cdot t^{2}$

$v_{o} = \frac{y-y_{o}-\frac{1}{2}\cdot g\cdot t^{2} }{t\cdot \sin \theta}$

$v_{o} = \frac{0\,m-20\,m-\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (8\,s)^{2} }{(8\,s)\cdot \sin 60^{\circ}}$

$v_{o} \approx 42.410\,\frac{m}{s}$

The initial speed of the projectile is approximately 42.410 meters per second.

b) The vertical component of the velocity of the projectile is determine by differentiating (2) in time and substitute all known values:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

( $v_{o} \approx 42.410\,\frac{m}{s}$ , $\theta = 60^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t = 4\,s$ )

$v_{y} = -2.5\,\frac{m}{s}$

The horizontal component of the velocity of the projectile is:

$v_{x} = v_{o}\cdot \cos \theta$

$v_{x} = 21.205\,\frac{m}{s}$

And the speed of the projectile is determined by Pythagorean Theorem:

$v = \sqrt{v_{x}^{2}+v_{y}^{2}}$

$v = \sqrt{\left(21.205\,\frac{m}{s} \right)^{2}+\left(-2.5\,\frac{m}{s} \right)^{2}}$

$v \approx 21.352\,\frac{m}{s}$

The speed of the projectile after 4 seconds is approximately 21.352 meters per second.

c) By (1) we find the horizontal range of the projectile:

( $x_{o} = 0\,m$ , $v_{o} \approx 42.410\,\frac{m}{s}$ , $\theta = 60^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t = 8\,s$ )

$x = 0\,m + \left(42.410\,\frac{m}{s} \right)\cdot (8\,s)\cdot \cos 60^{\circ}$

$x = 169.64\,m$

The horizontal range of the projectile is 169.64 meters.

d) The projectile reaches its maximum height when velocity is zero. By (3) and knowing that $v_{y} = 0\,\frac{m}{s}$ , $v_{o} \approx 42.410\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the time associated with maximum height is:

$0\,\frac{m}{s} = \left(42.410\,\frac{m}{s}\right)\cdot \sin 60^{\circ}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot t$

$t = 3.745\,s$

And by (2) and knowing that $y_{o} = 20\,m$ , $v_{o} \approx 42.410\,\frac{m}{s}$ , $t = 3.745\,s$ , $\theta = 60^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the maximum height reached by the projectile is:

$y = 20\,m + \left(42.410\,\frac{m}{s} \right)\cdot (3.745\,s)\cdot \sin 60^{\circ}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (3.745\,s)^{2}$

$y = 68.775\,m$

The maximum height of the projectile is 68.775 meters.

e) If we know that $y_{o} = 20\,m$ , $v_{o} \approx 42.410\,\frac{m}{s}$ , $t = 8\,s$ , $\theta = 60^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the components of the speed are, respectively:

$v_{y} = \left(42.410\,\frac{m}{s}\right)\cdot \sin 60^{\circ} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (8\,s)$