A potato is launched vertically upward with an initial velocity of 100 ft/s from

Question

A potato is launched vertically upward with an initial velocity of 100 ft/s from a potato gun at the top of a 75 ft tall building. The distance, in feet, that the potato travels from the ground after t seconds is given by s(t) = 16t^2 + 100t + 75. Find the velocity of the potato (in ft/s) after 0.75 s and 5.5 s. velocity at 0.75 s ft/s velocity at 5.5 s ft/s Find the speed of the potato (in ft/s) at 0.75 s and 5.5 s. speed at 0.75 s ft/s speed at 5.5 s ft/s Determine the time (in s) when the potato reaches its maximum height. s Find the acceleration (in ft/s^2) of the potato at 0.75 s and 3.5 s. acceleration at 0.75 s ft/s^2 acceleration at 3.5 s ft/s^2 Determine how long (in s) the potato is in the air. () s Determine the velocity (in ft/s) of the potato upon hitting the ground. () ft/s

Explanation / Answer

s = -16t^2 + 100t + 75

Velocity = derivative of position
v = -32t + 100

Plug in 0.75, we get :
v = -32(0.75) + 100
= -24 + 100
= 76

Plug in 5.5 :
v = -32(5.5) + 100
v = -176 + 100
v = 76

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Speed :
This is just the magnitude of velocity at that time

So, speed at 0.75s : 76

Speed at 5.5 s : 76

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MAx ht :
This will happen when the velocity is 0
s = -16t^2 + 100t + 75
s'= -32t + 100 = 0
32t = 100
t = 100/32
t = 3.125 seconds

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Acceleration :
We need to derive s twice
We have v = -32t + 100

Derive again and we have :
a = -32, which is a constant

So, a is a constant value no matter what t is

-32 ft/s2
-32 ft/s^2

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Lets find the times when s = 0
-16t^2 + 100t + 75 = 0
Quadratic formula gives us the times as
t = -0.67673 and 6.9267

Obviously, it will be t = 6.9267 when it hits the ground

So, it in air for 6.9267 - 0
i.e 6.9267 seconds

To two decimal places, we have :
6.93 seconds

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v = -32t + 100
Plug in 6.9267 in here, we have :

v = -32(6.9267) + 100
v = -121.65 ft/sec

Now, if this does not work, it means, we were to use the 6.93 value
with which our answer would be :
-121.76

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