Question

Please help asap!!

Answer 1

Answer:

Explanation:

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

That would mean -h = $v(\frac{9.8}{2})-\frac{1}{2}(9.80)(\frac{9.8}{2})^{2}.Since the ball has stopped at t = 9.8 = g, then that would mean that the final velocity v = 0.-h = [tex](0)(\frac{9.8}{2})-(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}$

$h = -(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}$

$h = (\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}$

$h = (\frac{9.8}{2})(\frac{9.8}{2})(\frac{9.8}{2})$

$h = \frac{9.8^{3}}{2^{3}}$

$h = \frac{941.192}{8}$

$h = 117.649$

The height of the cannonball at $t = \frac{9.8}{2}$ should be 117.649 m

Hope this helps!

Answer 2

Answer:

$h = 67.5\,m$

Explanation:

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

Half position

$h = H -\frac{1}{2}\cdot g \cdot \left(\frac{t_{g}}{2} \right)^{2}$

Final position

$0 = H -\frac{1}{2}\cdot g \cdot t_{g}^{2}$

The instant when the cannonball hits the ground is:

$t_{g} = \sqrt{\frac{2\cdot H}{g} }$

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

$h = H -\frac{1}{8}\cdot g \cdot \left(\frac{2\cdot H}{g} \right)$

$h = H -\frac{1}{4}\cdot H$

$h = \frac{3}{4}H$

$h = 67.5\,m$