Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Worst worded question to date: A brick is thrown upward from the top of a building at an angle of 25degrees to the horizontal and with an initial velocity of 15m/s. If the brick is in flight for 3 seconds, how tall is the building?

Physics
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

It took me a little bit to assume that the 3 seconds in flight was a terminanting time, and not just a random interval :)
|dw:1346854922454:dw|
\[-4.9(3)^2+15(3)sin(25^o)+h_o=0\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Could I suggest something? @amistre64 sir
of course
only the velocity in vertical direction is of use here \[V_v= 15 \sin 25 \] now let's find the time it takes the velocity in vertical direction to get to 0 and the distance it covers|dw:1346855419638:dw| let the height be h1 and time be t1 here acceleration due to gravity=10 m/s^2 in downwards direction so -10 \[v-u=at\] \[0-V_v=-10*t_1\] we could find t_1 from here \[v^2-u^2=2as\] \[0-(V_V)^2=2*-10*h_1\] h1 from here now the remaining time it takes to reach ground 3-t1 using this we could find h2 building's height is h2-h1
for h2 \[s=ut+1/2 at^2\] \[h_2=0\times (3-t_1)+\frac 12 (-10) (3-t_1)^2\]
while i can understand the concepts you presented, it does seem like an awful lot of extra workings. I am a mathematician at heart, and by default very lazy :)
:) he he

Not the answer you are looking for?

Search for more explanations.

Ask your own question