Solving a Motion Problem with Direct Integration Under certa

Solving a Motion Problem with Direct Integration Under certain conditions, we can assume a free-falling object undergoes constant vertical acceleration of -32 feet per second per second. Suppose a ball is thrown from the roof of a house that is 30 feet above the ground with a vertical velocity of 50 feet per second.

Solution

v=u+gt

g=-32

u=50

v can be 20 if ball is going up and -20 if ball is going down

Case 1 Ball going up

v=20=50-32t

t=30/32 s

Case 2. Ball going down

v=-20=50-32t

32t=70

t=70/32=2.1875

We need to check if ball is above the ground at this time

s=ut+gt^2/2 is formula to find displacement of the ball

Plugging, u=20 m/s,g=-32 m/s^2,t=2.1875 s

We get

s=-32.8 feet ie below the ground

Hence the ball does not attain 20 ft/s while going down