Asked by Martha
Wiley jumps straight up in the air from the edge of a cliff so that when he comes down he falls off the cliff. The cliff is 40 feet off the ground. His height t seconds after he jumps is s= -16t2 + 12t + 40.What is the maximum height? What is his maximum velocity?
Answers
Answered by
Reiny
ds/dt = -32t + 12
= 0 for a max of s
32t = 12
t = 12/32 sec = 3/8 sec
for that time
s = -16(9/64) + 12(3/8) + 40
= 169/4 or 42.25 ft above the base of the cliff
when he hits the ground, s = 0
-16t^2 + 12t +40=0
divide by -4
4t^2 - 3t - 10 = 0
(t-2)(4t+5) = 0
t = 2 or t = -5/4, but t > 0
so he will hit the ground after 2 seconds
ds/dt = -32t+12
when t=2
ds/dt = -32(2) + 12 = -52 ft/sec
the negative indicates a downward direction, he is falling
= 0 for a max of s
32t = 12
t = 12/32 sec = 3/8 sec
for that time
s = -16(9/64) + 12(3/8) + 40
= 169/4 or 42.25 ft above the base of the cliff
when he hits the ground, s = 0
-16t^2 + 12t +40=0
divide by -4
4t^2 - 3t - 10 = 0
(t-2)(4t+5) = 0
t = 2 or t = -5/4, but t > 0
so he will hit the ground after 2 seconds
ds/dt = -32t+12
when t=2
ds/dt = -32(2) + 12 = -52 ft/sec
the negative indicates a downward direction, he is falling
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