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# CBSE Class 9 Answered

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, 08:02: PM
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.

where g = 9.8 m/sis acceleration due to gravity.

----------------------------------------

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

......................... (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

.....................(2)
velocity after (t-1) secod is g(t-1)

Hence distance travelled in last 1 s time duration is

...................(3)
By dividing eqn.(2) by eqn.(3) , we get

From above expression, we get the following quadratic expression ,

By solving above quadratic expression , we get t = 2.5 s

From eqn.(1) , we get h as

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Question #3

If the body is projected upward with speed u , then maximum height h reached by the body is given as

............................. (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 " v= u2 - 2 g h " , we get

.............................(5)
If we substitute h from eqn.(4) , above eqn.(5) becomes

hence we get u= 800 m/ s2

Then we get height h from eqn.(4) as

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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

Hence u = 39.2 m/s
Answered by Thiyagarajan K | 06 Jul, 2023, 09:44: AM

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