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.

An object is dropped from a tower, 1600 ft above the ground. The object's height above the ground t seconds after the fall is s(t)=1600-16(t^2). Determine the velocity and acceleration of the object the moment it reaches the ground.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

You'll need the time that the object will reach the ground, which can be obtained by solving for s(t) = 0: 1600 - 16t^2 = 0 1600 = 16t^2 100 = t^2 t = 10 Once you have this time, you can plug it into the velocity and acceleration function, which the first and second derivative of s(t), respectively. v(t) = s'(t) = -32t a(t) = -32 (as expected, free fall acceleration due to gravity is constant) Then v(10) = -320 ft/s And a(10) = -32 ft/s^2
First determine how long it takes for the object to hit the ground. Set s(t) =1600 - 16t^2 = 0 So t = 10s would be the time taken. Take the first derivative of s(t) to find the velocity-time relationship: s'(t) = -32t So s(10) = -320ft/s... This is your velocity. I think the acceleration at all points close to the earth is roughly constant, -9.81m/s^2 = -32ft/s^2
thanks!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

So velocity is first derivative?
yes

Not the answer you are looking for?

Search for more explanations.

Ask your own question