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, 20:02: 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.
![begin mathsize 14px style h space equals space u space t space minus space 1 half g space t squared end style](https://images.topperlearning.com/topper/tinymce/cache/b01226db992833c443a39fe5a0fa23a6.png)
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](https://images.topperlearning.com/topper/tinymce/cache/0dd8d3fb4044e8ff20ded24a010c6a9b.png)
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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](https://images.topperlearning.com/topper/tinymce/cache/5bf17f240e8d8578c2a140482005a5e4.png)
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](https://images.topperlearning.com/topper/tinymce/cache/1d58a8532f200951ea1b2c8723f25422.png)
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](https://images.topperlearning.com/topper/tinymce/cache/93cb29df83de388f7320a9d4076a66a4.png)
![begin mathsize 14px style 0.64 space h space equals space g space open parentheses t minus 1 half close parentheses end style](https://images.topperlearning.com/topper/tinymce/cache/5037920c33bdb56290918165ab352824.png)
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](https://images.topperlearning.com/topper/tinymce/cache/9ad739a81011e74087a72db469f2ba75.png)
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](https://images.topperlearning.com/topper/tinymce/cache/95a423603cb34faffa3a8bc51bfb2f66.png)
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](https://images.topperlearning.com/topper/tinymce/cache/d52109b5bb1e1c32e087e9116d86e529.png)
----------------------------------------------
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](https://images.topperlearning.com/topper/tinymce/cache/b9ff4d260c7c37757f2355df50c439af.png)
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](https://images.topperlearning.com/topper/tinymce/cache/f3ef1804127cb87bf486f078a326945f.png)
![begin mathsize 14px style 400 space equals space u squared minus space left parenthesis space g space h space right parenthesis end style](https://images.topperlearning.com/topper/tinymce/cache/7f9f7e32eafb43d299ab3d455a53eebc.png)
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](https://images.topperlearning.com/topper/tinymce/cache/84bfdce10c0a20baa29904ddc19d0977.png)
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](https://images.topperlearning.com/topper/tinymce/cache/9fc61fd987a8d50e8f9dbdc5a29e1c53.png)
-------------------------------------------------
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](https://images.topperlearning.com/topper/tinymce/cache/89d33643e012da1f1c1ba699baccbc32.png)
Hence u = 39.2 m/s
Answered by Thiyagarajan K | 06 Jul, 2023, 09:44: AM
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