A test tube is knocked off a tower at the top of a building that is 400 feet abo

Question

A test tube is knocked off a tower at the top of a building that is 400 feet above the ground. (For our purposes, we will assume that air resistance is negligible.) The test tube drops 16 t^2 feet in t seconds

seconds.

(a) Calculate the average velocity in the first two seconds of the fall.

(Enter answer as a decimal number. Use the convention that velocities downward are negative.)

feet per secondincorrect

(b) Calculate the average velocity in the last two seconds of the fall.

(Enter answer as a decimal number. Use the convention that velocities downward are negative.)

feet per secondincorrect

(c) Calculate the instantaneous velocity when the test tube lands.

feet per secondincorrect

Explanation / Answer

S=400ft

X=16t2

dx/dt=32t

A) Average velocity in first two seconds=16(2)^2/2 = 16×4/2=32 ft/s

B) Average velocity in last two seconds =

S=400ft ,g= 32.147ft/s^2 ,u=0

Using second equation of motion

S=ut+1/2 gt^2

400=1/2×32.147t^2 so t=4.99s=5s

Distance covered in 5s =16×5×5=400ft

Distance covered in 3s= 16×3×3=144ft

So distance covered in last Two seconds = 256 ft

Average velocity in last two seconds wil 256/2=128ft/s

C)Instantaneous velocity =32t =32×5=160Ft/s

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