Respuesta :

First, determine how much of that 49m/s applies towards vertical velocity. So just multiply by the sine of 30. Giving
sin(30) * 49 m/s = 0.5 * 49 m/s = 24.5 m/s

Now the arrow will continue to climb until it's vertical velocity reaches 0. And since the attraction due to gravity is 9.8 m/s^2, that will be
24.5 m/s / 9.8 m/s^2 =  2.5 s

Finally, the distance an object travels under constant acceleration is given by the equation
d = 0.5 A T ^2
where
d = distance
A = acceleration
T = time.

So substitute the known values into the equation and solve for d.
d = 0.5 9.8 m/s^2 (2.5 s)^2
d = 4.9 m/s^2 * 6.25 s^2
d =  30.625 m
snehashish65

The height achieved by the arrow is of 30.62 m.

Given data:

The angle of inclination with Horizontal is, [tex]\theta =30^{\circ}[/tex].

The initial velocity of arrow is, u = 49 m/s.

Since, the inclination is given with respect to the horizontal, then the horizontal component of initial velocity is,

[tex]u' = u \times sin \theta\\\\u' = 49\times sin30\\\\u'=24.5 \;\rm m/s[/tex]

So, using the alternate expression for the horizontal component of initial velocity as,

[tex]g = \dfrac{u'}{t}[/tex]

Here, g is the gravitational acceleration.

Solving as,

[tex]9.8 = \dfrac{24.5}{t}\\\\\\t = \dfrac{24.5}{9.8}\\\\\\t = 2.5 \;\rm s[/tex]

Now, we can use the second kinematic equation of motion to find the height achieved by the arrow as,

[tex]h = u''t+\dfrac{1}{2}gt^{2}\\\\h = (0 \times t) +\dfrac{1}{2} \times 9.8 \times 2.5^{2}\\\\h = 30.62 \;\rm m[/tex]

Thus, we can conclude that the height achieved by the arrow is of 30.62 m.

Learn more about the kinematic equation of motion here:

https://brainly.com/question/14355103

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