Using calculus, h(t) = 128+16t−16t2 During what time interval will the ball be at least 32 ft above the ground?

Question

Using calculus, it can be shown that if a ball is thrown upwards with an initial velocity of 16 ft/s from the top of a building 128 ft high, then its height h above the ground t seconds later will be

h(t) = 128+16t−16t²

During what time interval will the ball be at least 32 ft above the ground?

Answer 1

If a ball is thrown upward, the formula is:

1/2 gt^2 + V0 t + X0 in here the V0 = 16 ft/s and the X0 = 128 ft (top of the building) and g = - 32 ft/s^2, so the function will be h(t) = -16t^2 + 16 t + 128 which is also given in the problem. We want the height be at least 32 ft, so:

-16 t^2 + 16 t + 128 > or = 32 simplify and get the roots

t^2 - t -6 < or = 0 (t -3)(t + 2) < or = 0 and the roots are t = 3 and -2 seconds (-2 means 2 seconds before)

Therefore the time interval that the ball will be at least 32 ft from the ground will be (0 , 3) starting from t = 0.

Answer 2

h(t).>=32

128+16t−16t^2>=32

96+16t−16t^2>=0

6+t-t^2>=0

the solution is -2<=t<=3