Question

A body thrown vertically upwards from the top of the building reaches the foot of the building in 5sec; but it will take 2sec if thrown vertically down with the same velocity.Calculate the height of the building.Given g = 9.8m/s^2.​

Answer 1

$\huge{\bold{\fcolorbox{black}{lime}{QUESTION:-}}}$$<font color = black >$

A body thrown vertically upwards from the top of the building reaches the foot of the building in 5sec; but it will take 2sec if thrown vertically down with the same velocity.Calculate the height of the building.Given g = 9.8m/s^2.

$\huge{\bold{\fcolorbox{black}{lime}{ANSWER:-}}}$$<font color = blue >$49m is the height.

$\huge{\bold{\fcolorbox{black}{lime}{Explanation:-}}}$$<font color = blue >$

When the object is thrown vertically upwards:

Initial velocity= -u ; a = g ; distance covered = h and t1 = time taken by the object to reach the foot of the building.

h = (-u)t1 × 1/2 gt^2

ut1 = 1/2 gt^2 - h ......... Equation (1)

When the object is thrown vertically downwards:-

Initial velocity= -u ; a = g ; distance covered = h and t2 = time taken by the object to reach the foot of the building.

Initial velocity= -u ; a = g ; distance covered = h and t2 = time taken by the object to reach the foot of the building.h = (-u)t1 × 1/2 gt^2base 2

ut1 = 1/2 gt^2 base 2 - h ......... Equation (2)

ut1 = 1/2 gt^2 base 2 - h ......... Equation (2)Dividing the equation (1) and (2), we get

t1 = 1/2 gt^2 base 1 - h

t2 = h - 1/2 gt^2 base 2

5 = 1/2 × 9.8 × (5)^2 - h

2 = h - 1/2 × (9.8)^2 × 2

5 = 122.5 - h

2 = h - 19.6

2 = h - 19.6

2 = h - 19.6

.: We get h = 49m

Thus, the height of the building is 49m.