Question

An object is dropped from a​ tower, 400 ft above the ground. The​ object's height above ground t seconds after the fall is ​s(t)equals400 minus 16 t squared. Determine the velocity and acceleration of the object the moment it reaches the ground. The velocity of the object the moment it reaches the ground is nothing ​ft/s.

Answer 1

Answer: v= 160ft/s

a=32ft/s^2 constant

Explanation:

s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.

0=400-16t^2

400=16t^2

25=t^2

t=5s

ds/dt=v(t)=0-32t

dv/dt=a(t)=-32 constant(gravity)

v(t)=-32(5s)= -160ft/s negative sign is only showing direction

Answer 2

The velocity of the object the moment it reaches the ground is 160 ft/s.

What is Velocity?

• Velocity is defined as the directional speed of an object.

How to solve this problem?

The problem can be solved by following steps.

s(t) = 400-16t^2 (given)

• We know that derivative of position is Velocity v(t) and the derivate of velocity is acceleration a(t)

Let s(t)=0

Therefore

0= 400-16t^2

400=16t^2

25=t^2

Therefore

t = 5sec

Now as we know that the derivative of the position is Velocity

so v(t) = ds/dt = -32t

where t = 5sec

substitute the value of t in v(t)

Therefore, v(t) = -32(5) = -160

The direction is negative

Hence the velocity is 160ft/s

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