Question

A body dropped from the top of a tower 200ft . heigh has an initial speed of 25ft/sec. What will be velocity of the body after 1 second of fall?
A.52 ft/sec
B. 57(U/sec
C. 62 ft/sec
D.48ft/sec​

Answer 1

Answer:

Answer is :  (B) 57(U/sec)

Explanation:For the velocity in the vertical (y) direction, the speed as a function of time v(t) is just

v(t)=v(0)-gt

We know v(0)=-25 fps,

v(t)=-25–32.2t

Therefore,

v(1)=-25–32.2(1)=-25–32.2=-57.2 fps

Solution: v(1)=-57.2 fps

Explanation:

Answer 2

Answer:

57(U/sec)

Explanation:

h=200ft

u=25ft/sec

v=?

g=9.8m/s²

Using the third equation of motion,i.e v²+u²=2gh

we have:-

v²+25²=2×9.8×200

v=√3295

v=57.402...(U/sec)

rounded off to 57(U/sec)