Question

a ball is thrown upward from the top of a 50 foot building with an initial velocity of 96 ft/sec. the position of the ball can be modeled by the following equation: h(t)=-16t^2+96t+50. At what point is the ball at its highest and determine the position of the ball at this time.

Answer 1

I have no idea what to do =(

Answer 2

Usually these kinds of problems is solved using calculus, but you marked this as precalculus, so extra information is needed. The gravitational acceleration on earth is about 32 ft/s^2, this means that for a ball moving upwards with a speed 96 ft/sec it will slow down with 32ft/s each second. So after 1 second it will be moving 96-32=64 ft/sec and after 2 seconds it will be moving 64-32=32 ft/sec, the third second it will be moving 32-32=0 ft/sec, or in otherwords not be moving at all. This means that the ball will start moving down right after this moment, so it must be at its highest. This is the solution, at t=3 seconds the ball will be at its highest and the position is given by the formula
\[h(3)=-16\times3^2+96\times3+50=194 ft\]

Answer 3

ok I understand
thank you

Answer 4

so 194 is the answer?