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anonymous
 5 years ago
An object is dropped from a tower, 1600 ft above the ground. The object's height above the ground t seconds after the fall is s(t)=160016(t^2). Determine the velocity and acceleration of the object the moment it reaches the ground.
anonymous
 5 years ago
An object is dropped from a tower, 1600 ft above the ground. The object's height above the ground t seconds after the fall is s(t)=160016(t^2). Determine the velocity and acceleration of the object the moment it reaches the ground.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'll need the time that the object will reach the ground, which can be obtained by solving for s(t) = 0: 1600  16t^2 = 0 1600 = 16t^2 100 = t^2 t = 10 Once you have this time, you can plug it into the velocity and acceleration function, which the first and second derivative of s(t), respectively. v(t) = s'(t) = 32t a(t) = 32 (as expected, free fall acceleration due to gravity is constant) Then v(10) = 320 ft/s And a(10) = 32 ft/s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First determine how long it takes for the object to hit the ground. Set s(t) =1600  16t^2 = 0 So t = 10s would be the time taken. Take the first derivative of s(t) to find the velocitytime relationship: s'(t) = 32t So s(10) = 320ft/s... This is your velocity. I think the acceleration at all points close to the earth is roughly constant, 9.81m/s^2 = 32ft/s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So velocity is first derivative?
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