A ball is thrown vertically upward with an initial velocity of $150 \mathrm{~m} / \mathrm{s}$. The ratio of velocity after 3 s and 5 s is $\frac{\mathrm{x}+1}{\mathrm{x}}$. The value of x is \_\_\_\_. Take ( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ).

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Kinematics

A ball is thrown vertically upward with an initial velocity of

150 m / s

. The ratio of velocity after 3 s and 5 s is

\frac{x + 1}{x}

. The value of x is ____.

Take (

g = 10 m / s^{2}

(A)6
(B)5✓ Correct
(C)-5
(D)10

Answer: (B)

Solution

Using the equation of motion

v = u - g t

(taking upward as positive):

At

t = 3

v (3) = 150 - 10 (3) = 150 - 30 = 120 m/s

t = 5

v (5) = 150 - 10 (5) = 150 - 50 = 100 m/s

Given ratio:

\frac{v ( 3 )}{v ( 5 )} = \frac{x + 1}{x}

\frac{120}{100} = \frac{x + 1}{x}

\frac{6}{5} = \frac{x + 1}{x}

6 x = 5 (x + 1)

6 x = 5 x + 5

x = 5

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option

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