(1 pt) If an arrow is shot straight upward on the moon with a velocity of 60 m/s, its height (in meters) after t seconds is given by s(t) = 60 t - 0.83 t^2. ?

What is the velocity of the arrow (include units) after 10 seconds?

How long will it take for the arrow to return and hit the moon (include units)?

With what velocity (include units) will the arrow hit the moon

1 Answer
Steve M
Jan 19, 2018

After #10s# the velocity is #43.4 \ ms^(-1) #

It will take #60/0.83 ~~ 72.289 \ s \ \ (3dp) # for the arrow to return back to ground level

The velocity when the arrow returns will be #-60 \ ms^(-1)#

Explanation:

We have:

# s(t) = 60t-0.83t^2 # ..... [A]

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

When #t=0# we have:

# s(t)=0+0 = 0 #

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

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

# v = (ds)/dt = 60 - (0.83)2t = 60 - 1.66t #

When #t = 10 => #

# v(10) = 60-1.66(10) = 60-16.6 =43.4 \ ms^(-1) #

We have ground level is at #s(0) \ ft#, so we can find the corresponding value of #t# corresponding to ground level:

# 60t-0.83t^2 =0 #

# :. t(60-0.83t)=0 #

Leading to:

# t = 0# or #60-0.83t=0 #
# => t=60/0.83 ~~ 72.289 \ s \ \ (3dp) #

We have already established that #t=0# corresponds to the time at which the arrow initially leaves the ground, so we can discard this solution. thus we have #t=60/0.83#

And we require the velocity when #t=60/0.83# corresponding to the arrow returning back to ground level. With # 60/0.83# we have:

# v = 60 - 2(0.83)(60/0.83) #
# \ \ = 60 - 120#
# \ \ = -60 \ ms^(-1)#

The velocity is negative because the ball is travelling vertically downwards

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