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HELP QUICK - An object is dropped from a​ tower, 159 ft above the ground. The​ object's height above ground t sec into the fall is s =159−16t^2. a. What is the​ object's velocity,​ speed, and acceleration at time​ t? b. About how long does it take the object to hit the​ ground? c. What is the​ object's velocity at the moment of​ impact?

Respuesta :

LammettHash
The velocity is the rate of change of the object's position:

[tex]s(t)=159-16t^2\implies v(t)=s'(t)=-32t[/tex]

Acceleration is the rate of change of velocity:

[tex]a(t)=v'(t)=s''(t)=-32[/tex]

The time it takes for the ball to hit the ground is the time [tex]t[/tex] for which [tex]s(t)=0[/tex], since [tex]s(t)[/tex] describes the position of the ball above the ground.

[tex]159-16t^2=0\implies 159=16t^2\implies t=\sqrt{\dfrac{159}{16}}\approx3.15[/tex]

(in seconds)

The object's velocity at this time can be found by plugging in this value of [tex]t[/tex] in to the velocity function.

[tex]v(3.15)\approx-32(3.15)\approx100.88[/tex]

(in feet per second)

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