dsouzaaww14
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Please help asap!!
Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?
Answer numerically in units of meters.

Please help asap Assume that the cannon is fired at time t0 and that the cannonball hits the ground at time tg What is the y position of the cannonball at the t class=

Respuesta :

Answer:

Explanation:

Let's use the equation Δy = [tex]v(t)-\frac{1}{2}(g)(t)^{2}[/tex]

That would mean -h = [tex]v(\frac{9.8}{2})-\frac{1}{2}(9.80)(\frac{9.8}{2})^{2}.

Since the ball has stopped at t = 9.8 = g, then that would mean that the final velocity v = 0.

-h = [tex](0)(\frac{9.8}{2})-(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]

[tex]h = -(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]

[tex]h = (\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]

[tex]h = (\frac{9.8}{2})(\frac{9.8}{2})(\frac{9.8}{2})[/tex]

[tex]h = \frac{9.8^{3}}{2^{3}}[/tex]

[tex]h = \frac{941.192}{8}[/tex]

[tex]h = 117.649[/tex]

The height of the cannonball at [tex]t = \frac{9.8}{2}[/tex] should be 117.649 m

Hope this helps!

xero099

Answer:

[tex]h = 67.5\,m[/tex]

Explanation:

The position of the cannonball is given by the following expressions:

Half position

[tex]h = H -\frac{1}{2}\cdot g \cdot \left(\frac{t_{g}}{2} \right)^{2}[/tex]

Final position

[tex]0 = H -\frac{1}{2}\cdot g \cdot t_{g}^{2}[/tex]

The instant when the cannonball hits the ground is:

[tex]t_{g} = \sqrt{\frac{2\cdot H}{g} }[/tex]

Lastly, this result is applied in the other equation, which simplified afterwards:

[tex]h = H -\frac{1}{8}\cdot g \cdot \left(\frac{2\cdot H}{g} \right)[/tex]

[tex]h = H -\frac{1}{4}\cdot H[/tex]

[tex]h = \frac{3}{4}H[/tex]

[tex]h = 67.5\,m[/tex]

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