anonymous
  • anonymous
a ball thrown vertically upwards from level ground is observed twice, at a height H above the ground, within a time interval t.The initial velocity of the ball was a. \[\sqrt{8gH+g ^{2}t ^{2}}\] b. \[\sqrt{8gH+\left( gT/2 \right)^{2}}\] c.\[1/2\sqrt{8gH+g ^{2}t ^{2}}\] d. \[\sqrt{8gH+4g ^{2}t ^{2}}\]
Physics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Here is one way to solve it. First we find the maximum height of the ball by the energy principle: \[mgh_{\max}= \frac{1}{2}mv_0^2 \Leftrightarrow h_{\max}= \frac{v_0^2}{2g}\] Then let's denote the distance the ball has dropped from its highest point (after time t/2) with d, and therefore \[d+H=h_{\max}\Leftrightarrow d=h_{max}-H\] Because the trajectory is symmetric, we can write \[d=\frac{1}{2}g(\frac{t}{2})^2\] and \[ h_{max}-H=\frac{1}{2}g(\frac{t}{2})^2 \] \[ \frac{v_0^2}{2g}-H=\frac{1}{2}g(\frac{t}{2})^2 \] from which we can solve \[v_0=\sqrt{\frac{1}{4}(gt)^2+2gH}=\frac{1}{2}\sqrt{(gt)^2+8gh}\] which is answer number c.
anonymous
  • anonymous
tnx a lot linjaaho !!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.