JEE Main 2021 (Online) 27th July Morning Shift | Motion in a Straight Line Question 64 | Physics

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $${h \over 3}$$ in both the directions.

$${{\sqrt 2 - 1} \over {\sqrt 2 + 1}}$$

$${1 \over 3}$$

$${{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$

$${{\sqrt 3 - 1} \over {\sqrt 3 + 1}}$$

JEE Main 2021 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

The relation between time t and distance x for a moving body is given as t = mx² + nx, where m and n are constants. The retardation of the motion is : (When v stands for velocity)

2 mv³

2 mnv³

2nv³

2n²v³

JEE Main 2021 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around :

(takes the value of g as 10 m/s²)

300 m

200 m

125 m

250 m

JEE Main 2021 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

The instantaneous velocity of a particle moving in a straight line is given as $$V = \alpha t + \beta {t^2}$$, where $$\alpha$$ and $$\beta$$ are constants. The distance travelled by the particle between 1s and 2s is :

3$$\alpha$$ + 7$$\beta$$

$${3 \over 2}\alpha + {7 \over 3}\beta $$

$${\alpha \over 2} + {\beta \over 3}$$

$${3 \over 2}\alpha + {7 \over 2}\beta $$

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