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

Asked by Benjamin | 19th Sep, 2015, 05:43: PM

Expert Answer:

Suppose that the object is thrown with velocity u from the top of the building of height h.
 
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
begin mathsize 12px style straight h space equals space open parentheses negative straight u close parentheses straight t subscript 1 space cross times space 1 half gt subscript 1 superscript 2 space ut subscript 1 space equals space 1 half gt subscript 1 superscript 2 space minus space straight h space........ space left parenthesis Equation space 1 right parenthesis end style
 
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
begin mathsize 12px style straight h space equals space ut subscript 2 space cross times space 1 half gt subscript 2 superscript 2 space ut subscript 2 space equals space straight h space minus space 1 half gt subscript 2 superscript 2 space space........ space left parenthesis Equation space 2 right parenthesis end style
Dividing equation (1) and (2), we get
 
begin mathsize 14px style straight t subscript 1 over straight t subscript 2 equals fraction numerator space 1 half gt subscript 1 superscript 2 space minus space straight h space over denominator straight h space minus space 1 half gt subscript 2 superscript 2 space space end fraction 5 over 2 equals fraction numerator space 1 half space cross times space 9.8 space cross times open parentheses 5 close parentheses squared space minus space straight h space over denominator straight h space minus space 1 half space cross times space 9.8 space cross times open parentheses 2 close parentheses squared space end fraction 5 over 2 equals fraction numerator space 122.5 space minus space straight h space over denominator straight h space minus space 19.6 end fraction On space solving comma space we space get straight h space equals space 49 space straight m end style
Thus, the height of the building is 49 m.

Answered by Yashvanti Jain | 20th Dec, 2017, 07:35: PM