A ball is thrown vertically upward from the top of a 96‐ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = –16t2 + 16t + 96. What is its velocity in ft/sec when it hits the ground?

1 Answer
Steve M
Jan 19, 2018

# -112 \ \ ft \ s^(-1)#

Explanation:

We have:

# s(t) = –16t^2 + 16t + 96 # ..... [A]

The question does not clarify if the displacement function, #s(t)# is relative to the height of the tower, the ground or in fact any other height.

When #t=0# we have:

# s(t)=0+0+96 = 96 #, which is the height of the tower.

Thus we can assert that the displacement, #s(t)# at time #t# is relative to ground level .

Knowing this, we have ground level is at #s(0) \ ft#, so we can find the corresponding value of #t# corresponding to ground level (and we expect one solution to be negative):

# –16t^2 + 16t + 96 = 0 #

# :. 16t^2 - 16t - 96 = 0 #

# :. t^2 - t - 6 = 0 #

# :. (t-3)(t+2) = 0 #

Leading to:

# t=-2# or # t=3#

#t=-2# corresponds to the time at which the ball would have been at ground level if it were thrown from ground level rather than at a height of #96 \ ft#, so we can discard this solution. thus we have #t=3#

Differentiating [A] wrt #t# we get the velocity function:

# v = (ds)/dt =-32t + 16 #

And we require the velocity when #t=3# corresponding to the ball being at ground level. With # t=3~ we have:

# v = (-32)(3) + 16 #
# \ \ = -96+16 #
# \ \ = -80 \ ft \ s^(-1)#

The velocity is negative because the ball is travelling vertically downwards

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