Question

A ball is being dropped from the top of a building so that its height above ground
after t seconds is (-I6t^2 + 94) feet. Find the velocity of the ball at the moment
when it is 13 feet above ground. A ball is being dropped from the top of a building so that its height above ground
after t seconds is (-I6t^2 + 94) feet. Find the velocity of the ball at the moment
when it is 13 feet above ground. @Mathematics

Answer 1

22m/s

Answer 2

it says that the answer is -72 ft/sec but I don't know how to get that answer

Answer 3

V(t)=d/dt{h(t)}

Answer 4

13=-16t2+94 is given from that you will get t as 9/4 sec put this time in above equation

Answer 5

h(t) = (-16t^2 + 94)
when h(t) = 13 we have: -16t^2 + 94 = 13
=> -16t^2 = -81
=> t^2 = 81/16 and hence t = 9/4 ( discard negative solution of t).
v(t) = h'(t) = -32t
Hence v(t = 9/4) is : -32(9/4) = -72.

Answer 6

yup

Answer 7

perfect, thanks for both your help!