Kinematics

JEE Main level, single correct, medium

46 of 308

Question 46 of 308

A ball is thrown vertically upward with an initial velocity of 150ms−1150m{s}^{-1}. The ratio of velocity after 3s3s and 5s5s is x+1x\dfrac{x+1}{x}. The value of xx is _____. {take, g=10ms−2g=10m{s}^{-2}}

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Answer: (D) 55

The formula to calculate the velocity of the ball after 3s3s is given by
v3=u−3g...(1){v}_{3}=u-3g...\left(1\right)
The formula to calculate the velocity of the ball after 5s5s is given by
v5=u−5g...(2){v}_{5}=u-5g...\left(2\right)
Divide equation (1) by equation (2) to obtain the required ratio.
v3v5=u−3gu−5g...(3)\dfrac{{v}_{3}}{{v}_{5}}=\dfrac{u-3g}{u-5g}...\left(3\right)
Substitute the values of the known parameters into equation (3) to calculate the required ratio.
v3v5=150−30150−50=120100=65=5+15\begin{matrix}\dfrac{{v}_{3}}{{v}_{5}} & = & \dfrac{150-30}{150-50} \\ & = & \dfrac{120}{100} \\ & = & \dfrac{6}{5} \\ & = & \dfrac{5+1}{5}\end{matrix}
Comparing the above equation with the given expression, it can be written that x=5x=5.

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