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question 1) A body is projected vertically upwards from top of the tower with the velocity of 30m/s. if it takes 8sec to reach on ground then what is the height of the tower (g=10ms). what is the total distance travelled by the body. question 2) A body is dropped from the height of h covers by 64% distance in its last second.find out thte total tim of the ball and height. question 3) A body projected vertically upward, At the half  of its maximum height its velocity becomes 20m/s, what is the maximum height attained by the body. question 5) A body is projected vertically upward from ground. the time of flight is 8 sec find the velocity of projection.
Asked by ketkiparasgandhi | 05 Jul, 2023, 20:02: PM
answered-by-expert Expert Answer
Question #1
 
If the body is projected  vertically upward from a height h with initial speed u = 30 m/s and it takes 8 s to reach the ground ,
then the vertical displacement of the body is calculated using the following equation of motion.
 
begin mathsize 14px style h space equals space u space t space minus space 1 half g space t squared end style
where g = 9.8 m/s2 is acceleration due to gravity.
 
 begin mathsize 14px style h space equals space left parenthesis space 30 space asterisk times space 8 space right parenthesis space minus space 1 half cross times 9.8 space cross times 8 space cross times 8 space equals space minus 73.6 space m end style
 
----------------------------------------
 
Question #2
 
If the body is dropped from a height h and it takes t seconds to reach ground ,
then the equation relating the vertical displacement h and time t is
 
begin mathsize 14px style h space equals space 1 half g space t squared end style ......................... (1)
It is given that 64% of total distance is travelled in last 1 s time duration of journey. 
Hence distance travelled in (t-1) seconds is 36% of total distance. 
 
Hence we have 
 
begin mathsize 14px style 0.36 space h space equals space 1 half g space left parenthesis t minus 1 right parenthesis squared end style  .....................(2)
velocity after (t-1) secod is g(t-1) 
 
Hence distance travelled in last 1 s time duration is 
 
begin mathsize 14px style 0.64 space h space equals space g left parenthesis t minus 1 right parenthesis space cross times 1 space plus space 1 half g space cross times 1 squared end style
begin mathsize 14px style 0.64 space h space equals space g space open parentheses t minus 1 half close parentheses end style...................(3)
By dividing eqn.(2) by eqn.(3) , we get
 
begin mathsize 14px style 9 over 16 space equals space fraction numerator left parenthesis t minus 1 right parenthesis squared over denominator left parenthesis 2 t minus 1 right parenthesis end fraction end style
 
From above expression, we get the following quadratic expression ,
 
 begin mathsize 14px style 16 t squared minus 50 t space plus 25 space equals space 0 end style
By solving above quadratic expression , we get t = 2.5 s 
 
From eqn.(1) , we get h as
 
begin mathsize 14px style h space equals space 1 half space g space t squared space equals space 1 half cross times 9.8 space cross times 2.5 cross times 2.5 space equals space 30.625 space m end style
----------------------------------------------
 
Question #3
 
If the body is projected upward with speed u , then maximum height h reached by the body is given as
 
begin mathsize 14px style h space equals space fraction numerator u squared over denominator 2 space g end fraction end style ............................. (4)
It is given that at half of the height velocity of the body is 20 m/s.
 
If we use this information in the equation of motion " v2 = u2 - 2 g h " , we get
 
begin mathsize 14px style 20 squared space equals space u squared space minus space 2 g space left parenthesis h divided by 2 right parenthesis end style
begin mathsize 14px style 400 space equals space u squared minus space left parenthesis space g space h space right parenthesis end style .............................(5) 
If we substitute h from eqn.(4) , above eqn.(5) becomes 
 
begin mathsize 14px style 400 space equals space u squared space minus space 1 half u squared end style
hence we get u2 = 800 m2 / s2
 
Then we get height h from eqn.(4) as
 
 begin mathsize 14px style h space equals space fraction numerator 800 over denominator left parenthesis 2 space cross times space 9.8 space right parenthesis end fraction space m space equals space 40.81 space m end style
-------------------------------------------------
 
Question #5 
 
when a body is projected upward with velocity u and it takes 8 s to reach the ground, 
 
then time taken to reach maximum height is 4 s.
 
Time t taken to reach maximum height is given by
 
begin mathsize 14px style t space equals space u over g space equals space fraction numerator u over denominator 9.8 end fraction space equals space 4 space s end style
 
Hence u = 39.2 m/s
Answered by Thiyagarajan K | 06 Jul, 2023, 09:44: AM

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